 Solution

A set is countable if it admits an injection into the natural numbers; this includes finite sets and the empty set. Use ${\mathbb{N}}_0=\{0,1,2,\dots \}$ for the enumeration. The Cantor pairing functionis a bijection ${\mathbb{N}}_0^2\to {\mathbb{N}}_0$. Indeed, on the diagonal $i+j=s$, its values are the consecutive integers from $s(s+1)/2$ through $s(s+1)/2+s$. These intervals are disjoint, and the next starts one beyond the preceding endpoint. Every nonnegative integer therefore determines exactly one diagonal $s$, then exactly one $j$, and finally $i=s-j$. If one's convention starts the natural numbers at one, translate both coordinates by one. Thus $\mathbb{N}\times \mathbb{N}$ is a countable set.

For countable sets $X,Y$, choose injections $e_X:X\to {\mathbb{N}}_0$ and $e_Y:Y\to {\mathbb{N}}_0$. The map $(x,y)\mapsto \Pi (e_X(x),e_Y(y))$ is an injection, proving their Cartesian product is countable. Repeating this proves that every finite Cartesian power of a countable set is countable.

For a sequence of countable sets $X_n$, choose an injection $e_n:X_n\to {\mathbb{N}}_0$ for each $n$. For $x\in {\bigcup }_n{X}_n$, let $n(x)$ be the least index containing $x$. The mapis an injection: its value recovers $n(x)$ and then $x$. This proves the countable union of countable sets theorem. Selecting all $e_n$ uses the usual axiom of countable choice; if enumerations are already supplied, the construction itself needs no additional choices.