Infinite Set has Countably Infinite Subset/Proof 2

Theorem
Every infinite set has a countably infinite subset.

Proof
Let $S$ be an infinite set.

First an injection $f: \N \to S$ is constructed.

In each non-empty subset $A \subset S$, let us choose one element $x_A \in A$.

Then, we define $f$ as follows:

Let $f \left({1}\right) = x_S$.

Now, if we have already defined $f \left({1}\right), \ldots, f \left({n}\right)$, we write:
 * $A_n = S \setminus \left\{f \left({1}\right), \ldots, f \left({n}\right)\right\}$

Since $S$ is infinite, $A_n$ is infinite for each $n \in \N$.

Now define $f \left({n+1}\right) = x_{A_n}$.

This finishes the definition of $f$.

To show that $f$ is injective, let $m, n \in \N$, say $m < n$.

Then:
 * $f \left({m}\right) \in \left\{{f \left({1}\right), \ldots, f\left({n-1}\right)}\right\}$

but:
 * $f \left({n}\right) \in S \setminus \left\{{f \left({1}\right), \ldots, f \left({n-1}\right)}\right\}$

Hence $f \left({m}\right) \ne f \left({n}\right)$.

So, now that we have an injection from $\N$ to $S$, let $S_0 = f \left({\N}\right)$, that is, $S_0$ is the image of $f$.

From Injection to Image is Bijection, $f$ is injective, it is a bijection from $\N$ onto its image.

Thus $S_0$ is countable.

Another way to see that $S_0$ is countable is to write:
 * $S_0 = \left\{{f \left({1}\right), f \left({2}\right), \ldots}\right\}$

Realizing that $S_0 \subset S$ completes the proof.

Warning
Choosing an element $x_A$ for each $A \subset X$ requires the Axiom of Choice.