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.

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

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


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

g \left({S}\right) & : n = 0 \\ g \left({S \setminus \left\{{f \left({0}\right), \ldots, f \left({n - 1} \right)}\right\} }\right) & : n > 0 \end{cases}$

Since $S$ is infinite, $S \setminus \left\{{f \left({0}\right), \ldots, f \left({n - 1} \right)}\right\}$ is non-empty for each $n \in \N$, so $f \left({\N}\right)$ is infinite.

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

Then:
 * $f \left({m}\right) \in \left\{{f \left({0}\right), \ldots, f\left({n-1}\right)}\right\}$

but:
 * $f \left({n}\right) \in S \setminus \left\{{f \left({0}\right), \ldots, f \left({n-1}\right)}\right\}$

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

Since $f$ is injective, it follows from Injection to Image is Bijection that $f$ is a bijection from $\N$ to $f \left({\N}\right)$.

Thus $f \left({\N}\right)$ is a countable subset of $S$.