Countable Union of Countable Sets is Countable/Proof 1

Theorem
Assuming the axiom of countable choice, a countable union of countable sets is countable.

Proof
Let $\left\langle{S_n}\right\rangle_{n \in \N}$ be a sequence of countable sets.

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

For all $n \in \N$, let $\mathcal F_n$ denote the set of all injections from $S_n$ to $\N$.

Since $S_n$ is countable, $\mathcal F_n$ is non-empty.

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

Let $\phi : S \to \N \times \N$, where $\times$ denotes the cartesian product, be the injection defined by:
 * $\phi \left({x}\right) = \left({n, f_n \left({x}\right)}\right)$

where $n$ is the least natural number such that $x \in S_n$.

Such a least $n$ exists because $\N$ is well-ordered.

There exists an injection $\alpha : \N \times \N \to \N$ because $\N \times \N$ is countable.

Then $\alpha \circ \phi: S \to \N$ is an injection because the composition of injections is an injection.

That is, $S$ is countable.