TITLE: How is axiom of choice utilized within the given proof?
QUESTION [5 upvotes]: I (only a beginner in set theory...) want to see a prove that an infinite set must have an infinite countable subset U:
The following answer sounds quite logically for me:
John Wayland Bales, Prove that every infinite set has a countable subset
However, I guess that this proof depends at least on on some weaker variant of axiom of choice.
John Wayland Bales constructs a series of non empty finite sets with increasing number of elements:

For each $N\in\mathbb{N}$, $S$ contains an element distinct from each
element in $U_N=\{x_1,x_2,\cdots,x_n\}$, so define
$U_{N+1}=U_N\cup\{x_{N+1}\}$ where $x_{N+1}$ is an element of $S$
distinct from each element of $U_N$.

Where is AC hidden in this argumentation?
I feel, that the last step is critical:

Let $$U=\bigcup_{N\in\mathbb{N}}U_N$$
Then $U$ is a countable subset of $S$.

I would like to learn, how the proof is done "correctly" by using AC or some weaker weaker variant (axiom of countable choice...).
My problem: Of course I see that recursive choices of "any" elements to create a new set are crucial, but those choices are done sequentially and not on an infinite set of sets that is already "available" and clearly defined. AC (in it's original textbook variant) is about the latter situation and not about a set, that is still in process of being constructed.

REPLY [3 votes]: First of all, a set $S$ is infinite iff it can't be put in bijection with $\varnothing$ or $I_n=\{1,2,...,n\} \ \ \forall n \in \mathbb{N}$. This implies that if $S$ is infinite ,it is not empty.
Formally speaking the recursion(where $x_1$ is a fixed element of $S$ that exists since $S\neq \varnothing$):
$$
\left\{
\begin{array}{c}
U_{1}=\{x_1\} \\
U_{n+1}=U_n \cup \{x_{n+1}\} \ \ \ x_{n+1}\in S/U_{n}
\end{array}
\right.
$$
Is ill-posed because it doesn't specify what is $x_{n+1}$.For example if $S=\mathbb{R}$ and $U_1=\{\pi\}$, who is $U_2$? It is not univocally determined , it may be $\{\pi,e\}$, it may be $\{\pi,2\}$,...
So this recursion is ill posed. To make this recursion well-posed we need something that takes "in input" $S/U_{n}$ and gives us an element of $S/U_{n}$ univocally for all $n$. The idea is to construct a choice function:
$$\Gamma: \mathcal{P}(S)/\{\varnothing \} \to S
$$
So that we can define the recursion as:
$$
\left\{
\begin{array}{c}
U_{1}=\{x_1\} \\
U_{n+1}=U_n \cup \{\Gamma (S/U_n)\} 
\end{array}
\right.
$$
Notice that this recursion is well posed since $S/U_n$ is non-empty for all $n$, so it can be an argument of choice function. Indeed if it was $S/U_n=\varnothing$ then $S=U_n$, but $S$ is infinite so $S\neq U_n$ (contradiction!).