Proof of A Countable Union of Countable Sets is Countable

Write $U=\bigcup_{m\in\mathbb{N}}X_{m}$, a subset of $S$. If $U=\emptyset$, it is countable and there is nothing to prove, so assume $U\ne\emptyset$ and fix $z\in U$.

Step 1: each $X_{m}\cup\{z\}$ is nonempty and countable. Fix $m\in\mathbb{N}$. Since $X_{m}\subseteq S$ is countable and $z\in S$, claim 6 of Basic Properties of Countable Sets shows that $X_{m}\cup\{z\}$ is countable; it contains $z$, so it is nonempty. Consequently some sequence has $X_{m}\cup\{z\}$ as its set of terms, and since $X_{m}\cup\{z\}\subseteq U$, that sequence is a sequence in $U$.

Step 2: a simultaneous choice of enumerations. As in the definition of the set of tuples in a set, where the maps from an initial segment of $\mathbb{N}$ to a set are collected into a set, we use that the maps from one set to another form a set. Accordingly, let $\Sigma$ be the set of all sequences in $U$, that is, of all families in $U$ indexed by $\mathbb{N}$, and, for $m\in\mathbb{N}$, let

$$A_{m}=\bigl\{s\in\Sigma:\ \{s_{n}:n\in\mathbb{N}\}=X_{m}\cup\{z\}\bigr\}.$$

By step 1, $A_{m}$ is nonempty for every $m$, and $(A_{m})_{m\in\mathbb{N}}$ is a family of subsets of $\Sigma$. By Axiom of Countable Choice there is a sequence $(a_{m})_{m\in\mathbb{N}}$ in $\Sigma$ with $a_{m}\in A_{m}$ for every $m\in\mathbb{N}$. Write $a_{m,n}$ for the $n$-th term of the sequence $a_{m}$.

Step 3: conclusion. Define

$$g:\mathbb{N}\times\mathbb{N}\to U,\qquad g(m,n)=a_{m,n},$$

which indeed takes values in $U$ because $a_{m}$ is a sequence in $U$. If $u\in U$, then $u\in X_{m}$ for some $m\in\mathbb{N}$, hence $u\in X_{m}\cup\{z\}$, which is the set of terms of $a_{m}$; so $u=a_{m,n}=g(m,n)$ for some $n\in\mathbb{N}$. Therefore $U=\{g(w):w\in\mathbb{N}\times\mathbb{N}\}$. Since $\mathbb{N}\times\mathbb{N}$ is countable by The Set of Pairs of Natural Numbers is Countable, claim 4 of Basic Properties of Countable Sets shows that $U$ is countable.