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 $T$.

Then $\mathcal F_n$ is non-empty because an infinite set has subsets of any finite cardinality.

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

Let $\displaystyle T = \bigcup_{n \in \N} S_n$. Then $T \subseteq S$.

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

Hence $T$ is infinite.

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