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:
 * $\FF_n = \set {T \subseteq S: \size T = n}$

where $\size T$ denotes the cardinality of $T$.

From Set is Infinite iff exist Subsets of all Finite Cardinalities:
 * $\FF_n$ is non-empty.

Using the axiom of countable choice, we can obtain a sequence $\sequence {S_n}_{n \mathop \in \N}$ such that $S_n \in \FF_n$ for all $n \in \N$.

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

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

From Set is Infinite iff exist Subsets of all Finite Cardinalities:
 * $T$ is infinite.

From Countable Union of Countable Sets is Countable, $T$ is countable.