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.). 

Let $n \in \mathbb{N}$. Note $n$ is finite.
Let $A_n \subseteq \mathbb{N}$ where $n = max(A_n)$.
Since $n$ is finite, $A_n$ is finite.
$\mid \mathcal{P}(A_n) \mid = 2^n$. Since $n$ is finite, $2^n$ is finite.
$\mathcal{F}(\mathbb{N}) = \bigcup_{n=1}^\infty \mathcal{P}(A_n)$
Since union of countable sets is countable and $\mathcal{F}(\mathbb{N})$ is such, it is countable.
