Infinite Set has Countably Infinite Subset/Proof 2

Proof
Let $S$ be an infinite set.

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

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

Then define $f: \N \to S$ as follows:


 * $\forall n \in \N: f \left({n}\right) = \begin{cases}

\map g S & : n = 0 \\ \map g {S \setminus \set {\map f 0, \ldots, \map f {n - 1} } } & : n > 0 \end{cases}$

Since $S$ is infinite, $S \setminus \set {\map f 0, \ldots, \map f {n - 1} }$ is non-empty for each $n \in \N$.

Therefore $f \sqbrk \N$ is infinite.

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

Then:
 * $\map f m \in \set {\map f 0, \ldots, \map f {n - 1} }$

but:
 * $\map f n \in S \setminus \set {\map f 0, \ldots, \map f {n - 1} }$

Hence $\map f m \ne \map f n$.

Since $f$ is injective, it follows from Injection to Image is Bijection that $f$ is a bijection from $\N$ to $f \sqbrk \N$.

Thus $f \sqbrk \N$ is a countable subset of $S$.