Neatest proof that set of finite subsets is countable? I am looking for a beautiful way of showing the following basic result in elementary set theory:

If $A$ is a countable set then the set of finite subsets of $A$ is
  countable.

I proved it as follows but my proof is somewhat fugly so I was wondering if there is a neat way of showing it:
Let $|A| \le \aleph_0$. If $A$ is finite then $P(A)$ is finite and hence countable. If $|A| = \aleph_0$ then there is a bijection $A \to \omega$ so that we may assume that we are talking about finite subsets of $\omega$ from now on. Define a map $\varphi: [A]^{<\aleph_0} \to (0,1) \cap \mathbb Q$ as $B \mapsto \sum_{n \in \omega} \frac{\chi_B (n)}{2^n}$. Then $\varphi$ is injective hence the claim follows. (The proof of which is what is the core-fugly part of the proof and I omit it.). 

A proof for finite subsets of $\mathbb{N}$:
For every $n \in \mathbb{N}$, there are finitely many finite sets $S \subseteq \mathbb{N}$ whose sum $\sum S = n$.
Then we can enumerate every finite set by enumerating all $n \in \mathbb{N}$ and then enumerating every (finitely many) set $S$ whose sum is $\sum S = n$.
Since every finite set has such an $n$, every finite set is enumerated. QED.
If you want the proof to hold for any countable $A$, first define any injective function $f: A \to \mathbb{N}$.
