Infinite Set has Countably Infinite Subset/Proof 3

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.

Let $f$ be a choice function on $\mathcal P \left({S}\right) \setminus \left\{{\varnothing}\right\}$.

That is:
 * $\forall A \in \mathcal P \left({S}\right) \setminus \left\{{\varnothing}\right\}: f \left({A}\right) \in A$

This is justified only if the Axiom of Choice is accepted.

Let $\mathcal C$ be the set of all finite subsets of $S$.

Let $A \in \mathcal C$.

Since $S$ is infinite it follows that $S \setminus A \ne \varnothing$.

So $S \setminus A \in \operatorname{Dom} \left({f}\right)$.

Let $g: \mathcal C \to \mathcal C$ be the mapping defined as:
 * $g \left({A}\right) = A \cup \left\{{f \left({S \setminus A}\right)}\right\}$

That is, $g \left({A}\right)$ is constructed by joining $A$ with the element that $f$ chooses from $S \setminus A$.

Consider the Recursion Theorem applied to $g$, starting with the set $\varnothing$.

We obtain a mapping $U: \N \to \mathcal C$ such that:
 * $U \left({x}\right) = \begin{cases}

\varnothing & : x = 0 \\ U \left({n}\right) \cup \left\{{f \left({S \setminus U \left({n}\right)}\right)}\right\} & : x = n^+ \end{cases}$ where here $\N$ is considered as elements of the minimal infinite successor set $\omega$.

Consider the mapping $v: \N \to S$, defined as:
 * $\forall n \in \N: v \left({n}\right) = f \left({S \setminus U \left({n}\right)}\right)$

We have that, by definition of $v$:
 * $(1): \quad \forall n \in \N: v \left({n}\right) \notin U \left({n}\right)$


 * $(2): \quad \forall n \in \N: v \left({n}\right) \in U \left({n^+}\right)$


 * $(3): \quad \forall m, n \in \N: n \le m \implies U \left({n}\right) \subseteq U \left({m}\right)$

Then because $v \left({n}\right) \in U \left({m}\right)$ but $v \left({m}\right) \notin U \left({m}\right)$:


 * $(4): \quad \forall m, n \in \N: n < m \implies v \left({n}\right) \ne v \left({m}\right)$

So $(4)$ implies that $v$ maps distinct elements of $\N$ onto distinct elements of $S$.

Thus $v: \N \to S$ is an injection.

It follows from Injection to Image is Bijection that $v$ is a bijection from $\N$ to $v \left({\N}\right)$.

Thus $v \left({\N}\right)$ is the countable subset of $S$ that was required.