Infinite Set has Countably Infinite Subset/Proof 3

Proof
Let $S$ be an infinite set.

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

Let $f$ be a choice function on $\powerset S \setminus \set \O$.

That is:
 * $\forall A \in \powerset S \setminus \set \O: \map f A \in A$

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

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

Let $A \in \CC$.

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

So $S \setminus A \in \Dom f$.

Let $g: \CC \to \CC$ be the mapping defined as:
 * $\map g A = A \cup \set {\map f {S \setminus A} }$

That is, $\map g A$ 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 $\O$.

We obtain a mapping $U: \N \to \CC$ such that:
 * $\map U x = \begin{cases}

\O & : x = 0 \\ \map U n \cup \set {\map f {S \setminus \map U n} } & : x = n^+ \end{cases}$

where here $\N$ is considered as the set of elements of the von Neumann construction of natural numbers.

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

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


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


 * $(3): \quad \forall m, n \in \N: n \le m \implies \map U n \subseteq \map U m$

Then because $\map v n \in \map U m$ but $\map v m \notin \map U m$:


 * $(4): \quad \forall m, n \in \N: n < m \implies \map v n \ne \map v m$

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 $\map v \N$.

Thus $\map v \N$ is the countable subset of $S$ that was required.