TITLE: (Long) Detailed Proof of Kőnig's Lemma (Explicit, Down to Axiom of Choice)
QUESTION [10 upvotes]: Kőnig's Lemma states that in an infinite, locally finite, connected graph $G$, there exists an infinite one-way path (a ray). The proof of it in my graph theory book (Introduction to Graph Theory, 4th ed., Wilson) had a sour taste to it, like it was covering something up. So I checked out proofs from other sources, which also seemed to cover up something which felt very close to the foundation (i.e. using the axioms). After looking it up I found out that this is indeed the case, as it results directly from the axiom of dependent choice (DC).
This wasn't a question on an assignment for me or anything, but I wanted to try to give a more detailed, albeit lengthy proof of the lemma with more explicit constructions and play-by-play, highlighting some of the key points and bare bones math going on here (including the general statement of the lemma and the use of DC), to the best of my understanding. I was hoping that anyone with time to spare hands could review or give comment, make sure everything I'm stating things correctly! And of course, I would be happy if this helps anyone in a similar predicament that comes to read this in the future, or helps anyone to appreciate the deeper levels of a simple theorem. 
I've enabled the "answer your own question" box so I can post my proof separately, but I encourage/challenge others to do the same and make a detailed construction as an exercise, especially during these times of isolation!
Below I'll prove a similar theorem which uses Kőnig's Lemma in a more general form (but sweeps DC under the rug). It's a good reference because it's easier to understand without a super explicit construction:

Let $G$ be a countable graph such that every finite subgraph of $G$ is $k$-colourable. Then $G$ is $k$-colourable.

Proof 
Since $G$ is countable, its vertices are enumerable as $v_1$, $v_2$, $v_3$, etc. Let $G_n$ be the (finite) subgraph induced by vertices $v_0$ through $v_n$. Since each vertex $v_n$ corresponds to a finite induced subgraph $G_n$, there are countably many $G_n$. Since $G_n \subset G_{n+1}$ for all $n$ by construction, the union over all $G_n$ yields $G$. Since each $G_n$ is finite, there are countably many corresponding sets $C_n$ of valid $k$-colourings of $G_n$, each non-empty with a finite number of elements given by the chromatic polynomial of $G_n$, $P_{G_n}(k)$. If we consider a colouring of $G_{n+1}$ and remove vertex $v_{n+1}$, we are left with a valid colouring of $G_n$: removing vertices does not invalidate a colouring. And so this valid colouring of $G_n$ must be an element of $C_n$. In general, for every colouring $c_{n+1} \in C_{n+1}$ of $G_{n+1}$ there is some colouring $c_{ind} \in C_{n}$ such that $c_{ind} \prec c_{n+1}$ (i.e. $c_{ind}$ is induced on $G_n$ by $c_{n+1}$). Kőnig's Lemma states that since there are countably many non-empty $C_n$ that follow these conditions, we must have a countable set of $c_n \in C_n$ such that $c_n \prec c_{n+1}$ for all $n$. If there were not, then all inductive sequences of colourings would terminate at some finite point $t$. If we take $\tau$ to be the maximum such $t$ over all sequences, then $C_{\tau + 1}$ must be empty, contradictory to our assumption that all $G_n$ are $k$-colourable (and in turn that all $C_n$ are non-empty). Therefore there is a countable set of inductive valid $k$-colourings $c_n$, and their union gives us a valid $k$-colouring of $G$. 
Thus, $G$ is $k$-colourable. $\blacksquare$ 
The general form of Kőnig's Lemma can be stated as follows (paraphrasing Infinite Graphs - A Survey, by Nash-Williams, 1967):

Given a countable sequence of finite, non-empty, disjoint sets $S_n$ and a relation $\prec$ on $\bigcup S_n$, if for each element $s_{n+1} \in S_{n+1}$ there exists an $s_n \in S_n$ such that $s_n \prec s_{n+1}$, then there exists a countable sequence of elements $(s_n)$ such that $s_n \prec s_{n+1}$ for all $n$. 

The end of the $k$-colouring proof I gave is basically the proof of this theorem. In this general form, the $S_n$ can be thought of as "configuration spaces" (finite sets of valid configurations), and the relation $\prec$ can be thought of as an inductive consistency relation. In the $k$-colouring proof, our configuration spaces are the $C_n$ and they are finite due to the chromatic polynomial of $G_n$, though I didn't explicitly construct them so we can't show that they are disjoint. But as I mentioned, this is still easier to understand without explicit construction. This is because we iterate out $G$ vertex by vertex, and as a result, we get a natural correspondence between the iterated subgraphs and their colourings. In proving the graph theoretic result about the existence of an infinite one-way path, the iteration scheme is not vertex by vertex, and so more explicit construction is needed to make the simple argument as above. That will be in my proof below, and similar methods can be used to construct a more explicit proof of this $k$-colourability result, and one for planarity as well!

REPLY [4 votes]: Dependent Choice is quite significantly stronger than what is necessary to prove Kőnig's lemma. And although the "obvious proof" uses Dependent Choice, we can do with much less. Indeed, Kőnig's lemma is equivalent to the statement "Every countable family of finite sets admits a choice function".
This principle is very weak. It follows from "the countable union of finite sets is countable", which follows from "the countable union of countable sets is countable", which follows from "every countable family of sets admits a choice function", which follows from Dependent Choice. None of these implications is reversible, either.
In fact, Kőnig's lemma follows even from "every set can be linearly ordered", or "every infinite set contains a countably infinite subset".
Suppose that $G$ is a locally finite graph. We may assume that it is cycle-free, otherwise pick any cycle, and loop through it infinitely many times. Now that the graph is cycle-free, we may also assume it is connected, otherwise we can restrict to a connected component.
Fix a node, $g_0\in G$. Then we have a tree with $g_0$ being the root. Define $T_n$ to be the set of vertices for which the unique path from $g_0$ has exactly $n$ nodes. Because $G$ is cycle-free and connected, these paths are indeed unique, and every node in $G$ lies in some $T_n$.
Claim 1. For each $n$, $T_n$ is finite.
Proof. $T_0=\{g_0\}$. Suppose that $T_n$ is finite, then elements of $T_{n+1}$ are all neighbours of elements of $T_n$. But as $G$ is locally finite, $T_{n+1}$ is a subset of a finite union of finite sets, which means it is finite. $\square$
Say that a vertex $g$ in $G$ is terminal if there is some $\ell_g$ such that all paths going from $g_0$ through $g$ are at most of length $\ell_g$.
Claim 2. For each $n$, $T_n$ contains at least one node which is not terminal.
Proof. If all the nodes in $T_n$ are terminal, let $\ell = \max\{\ell_g\mid g\in T_n\}$, then $G=\bigcup_{k\leq \ell}T_k$, and so $G$ is finite. But we assumed that $G$ is infinite. $\square$
Claim 3. If $g\in T_n$ and $g$ is not terminal, then there is some $g'\in T_{n+1}$ such that $g'$ is a neighbour of $g$, and $g'$ is also not terminal.
Proof. If $g$ is connected only to terminal nodes in $T_{n+1}$, let $\ell$ be $\max\{\ell_{g'}\mid g'\in T_{n+1}\text{ and it is a neighbour of }g\}$. That means that every path from $g_0$ going through $g$ must have at most $\ell$ vertices, and so $g$ is terminal as well. $\square$
Let $T'_n$ denote the subset of $T_n$ of those nodes which are not terminal. Since $T'_n$ is finite, say with $k_n$ elements, the set $S_n=\{f\colon\{0,\dots,k_n-1\}\to T'_n\mid f\text{ is a bijection}\}$ is also finite, and non-empty. Let $F$ be a choice function from $\{S_n\mid n\in\Bbb N\}$.
Finally, define $f$ on the graph as follows: $$f(g)=\begin{cases}
F(n+1)(i) & g\in T_n, i=\min\{j<k_{n+1}\mid F(j)\text{ is a neighbour of }g\},\\
g & \text{otherwise}.\end{cases}$$
That is, $f$ either "stops" if there are only terminal nodes, or picks the neighbour with the least index in the enumeration of $T'_{n+1}$, if such neighbour exists.
And now I claim that applying the recursion theorem on $f$, with $g_0$ as a starting point, will produce a ray. That is, a sequence $s$ such that $s_0 = g_0$ and $s_{n+1} = f(s_n)$.
Claim 4. The sequence $s$ is a ray.
Proof. It is enough to claim that for all $n$, $s_n$ is not a terminal node. For $n=0$ we already know that $g_0$ is not terminal. But by Claim 3, if $s_n$ is not terminal, then there is some neighbour in $T_{n+1}$ which is also not terminal, so $s_{n+1}=f(s_n)$ must be such neighbour, and in particular it is not terminal. $\square$

You can see from this proof how the various choice principles I've pointed out allow us to prove the existence of the choice function $F$.

If the countable union of finite sets is countable, then we take the union of all the $T'_n$, it is countable, so we may enumerate it, and use this enumeration to define $F$.
If every set can be linearly ordered, we may linearly order the union of the $T'_n$s, which again provides us with a uniform enumeration, as a linear ordering of a finite set is a well-ordering.
If every infinite set has a countably infinite subset, then the union of the $T'_n$s has a countably infinite subset, it follows that it must meet infinitely many $T'_n$s, but by the uniqueness of the paths, we can then create a countable subtree, and use the enumeration, again.

Finally, to obtain the countable choice for families of finite sets axiom, let $\{A_n\mid n\in\Bbb N\}$ be a family of non-empty finite sets. We define $G_n$ to be the set of choice functions from $A_k$ for $k<n$, i.e. $\prod_{k<n}A_k$, and let $G=\bigcup_{n\in\Bbb N}G_n$, with the edge relation defined as follows: $\{f,g\}$ is an edge if and only if there is some $n$ such that $\operatorname{dom} g=\{0,\dots,n\}$ and $f=g\restriction\{0,\dots,n-1\}$, or vice versa.
That is, an edge is between two functions such that the one extends the other by a single new choice. It is easy to see this is locally finite, as each $A_n$ is finite. So there is a ray, from which we define a choice function from the entire family $\{A_n\mid n\in\Bbb N\}$.