Countable Union of Countable Sets is Countable/Proof 2

Theorem
Assuming the axiom of countable choice, the union of a countable number 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 $\mathcal F_n$ be the set of all surjections from $\N$ to $S_n$.

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 $\phi : \N \times \N \to S$, where $\times$ denotes the cartesian product, be the surjection defined by:
 * $\phi \left({m, n}\right) = f_m \left({n}\right)$

Let $\alpha : \N \to \N \times \N$ be a surjection, which exists because $\N \times \N$ is countable.

Then the mapping $f = \phi \circ \alpha$, where $\circ$ denotes composition, is a surjection from $\N$ to $S$ because the composition of surjections is a surjection.

The preimage (under $f$) of any two distinct elements of $S$ are non-empty and disjoint, so their minimal elements, which exist because $\N$ is well-ordered, are distinct.

Therefore, the mapping $F : S \to \N$ defined by:
 * $F \left({x}\right) = \min f^{-1} \left({x}\right)$

is an injection.