Infinite Set has Countably Infinite Subset/Proof 4

Theorem
If the axiom of countable choice is accepted, then it can be proven that every infinite set has a countably infinite subset.

Proof
Let $S$ be an infinite set.

For all $n \in \N$, let:
 * $\mathcal F_n = \left\{{T \subseteq S : \left\vert{T}\right\vert = n}\right\}$

where $\left\vert{T}\right\vert$ denotes the cardinality of $T$.

Because an infinite set has subsets of any finite cardinality, it follows that $\mathcal F_n$ is non-empty.

Using the axiom of countable choice, we can obtain a sequence $\left\langle{S_n}\right\rangle_{n \in \N}$ such that $S_n \in \mathcal F_n$ for all $n \in \N$.

Define:
 * $\displaystyle T = \bigcup_{n \mathop \in \N} S_n \subseteq S$

For all $n \in \N$, $S_n$ is a subset of $T$ whose cardinality is $n$.

Hence, it follows that $T$ is infinite.

Because the countable union of finite sets is countable, $T$ is countable.