Neatest proof that set of finite subsets is countable?

If $A$ is an infinite countable set, $S$ is the set of finite subsets of $A$ and $S_k$ is the set of $k$-element subsets of $A$, then

$$ |S| = \sum_{k \in \mathbb{N}} |S_k| \leq \sum_{k \in \mathbb{N}} |A|^k \leq \sum_{k \in \mathbb{N}} \aleph_0 = |\mathbb{N}| \cdot \aleph_0 = \aleph_0 $$

Subsets of $A$ are in bijection with subsets of $\mathbb{N}$ so it suffices to enumerate the latter. Any subset of $\mathbb{N}$ can be written as a finite string using the following thirteen characters $$0\ 1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9\ \{\ \}\ ,$$ By setting $${\{}=10 \qquad {\}}=11 \qquad {,}=12$$ writing out a subset of $\mathbb{N}$ is a base-$13$ expression of an integer. For example $$\{ 0,1 \}_{13} = 10 \cdot 13^4 + 0 \cdot 13^3 + 12 \cdot 13^2 + 1 \cdot 13 + 11 = 287662$$

The map sending $$S \mapsto \text{the least}\ n\ \text{represented by}\ S$$ defines an injection into $\mathbb{N}$.

It is sufficient to prove the claim for $A=\mathbb{N}$. Denote by $\operatorname{Fin}(\mathbb{N})$ the set of all finite subsets of $\mathbb N$. We have $$\operatorname{Fin}(\mathbb{N}) = \bigcup_{n\in\mathbb{N}}\left\{ A\subseteq\mathbb{N}: \max(A) = n \right\},$$ which is a countable union of finite sets as for each $n\in\mathbb{N}$ there certainly are less than $2^n = \left|\mathcal{P}(\{1,\ldots,n \})\right|$ subsets of $\mathbb{N}$ whose largest element is $n$. Hence, $\operatorname{Fin}(\mathbb{N})$ is countable itself.