TITLE: Every infinite set contains a countable subset
QUESTION [7 upvotes]: Assuming the axiom of choice, it can be shown that every infinite set contains a countable subset.
Also, it is my understanding that the proof of this statement requires at least countable choice - i.e., it is consistent with ZF that there exist infinite sets which contain no countable subset.
However, consider the following proof:
Given an infinite (i.e. not equinumerous to any natural number) set $S$, let $A_1$ be a set containing one element in $S$, and construct $A_n$ by adding to $A_{n-1}$ any element from $S$ not already in $A_{n-1}$. This can be done since $S$ is finite, so at no point is $A_{n-1}$ equal to all of $S$. We can construct every $A_n$ using only ZF, since we are only making finitely many choices. Now, define $f: \mathbb{N} \rightarrow S$ by defining $f(n)$ to be the unique element in $A_n \backslash A_{n-1}$. $f$ is injective since $f(n) \notin A_m$ for every $m < n$, so $f(n) \neq f(m)$. Hence, $f$ is an injection from $\mathbb{N}$ to $S$, and the image of $f$ is a countable subset of $S$.
My problem is figuring out where in this proof the axiom of choice is used. It seems the construction of the $A_n$'s can be done without AC, since we are only making a finite number of choices. The construction of $f$ involves choosing an element from $A_n \backslash A_{n-1}$, but since we have a systematic way for choosing (take the unique element in the set) that should be allowed. I know this proof must depend on the axiom of choice, but where is it used?

REPLY [3 votes]: I think it will clarify things to think about how you would actually set up the definition of the $A_n$ by induction.  Spelling it out in detail will show exactly where the axiom of choice is used.
First, the formal principle of definition by mathematical induction that you are using looks like this:

Principle of definition by mathematical induction:
To define a function $f$ from $\omega$ to a set $E$ by induction, start with a function $g\colon E \to E$ and a specific $e_0\in E.$ You can then conclude that there is a unique $f\colon\omega\to E$ such that for all $n\lt\omega,$
  $$f(n)=\begin{cases}e_0,&\text{ if }n=0,
\\ g(f(n-1)),&\text{ if }n\gt 0.
\end{cases}$$

(There are variants—for example, $g$ can be allowed to depend on $n$ also, or you could set things up for strong induction, but the above is sufficient for what you're doing.)
The function $f$ you are defining by induction is $n\mapsto A_n.$  To fit it into the above template, you can set $E$ equal to the collection of all finite subsets of $S$ (where $S$ is the infinite set that you are starting out with), and you can set $e_0=\emptyset$ (you actually started out with $A_1$ being a $1\text{-element}$ set, but this is easier).
But what are you going to use for $g?$
For your construction to work, the function $g\colon E\to E$ needs to satisfy the following property:

For any finite subset $X$ of $S,$ $g(X)\in S\setminus X.$ 

It's true that for any finite subset $X$ of $S,$ the set $S\setminus X$ is non-empty. Nevertheless, it requires a use of the axiom of choice to conclude that there is a function $g$ with domain $E$ such that for all $X$ in $E,$ $\;g(X)\in S\setminus X.$