countably infinite union of countably infinite sets is countable

The answer depends on your set theory.

If your set theory includes the Axiom of (Countable) Choice, then you can proceed as follows:

1. For each $n\in\mathbb{N}$, select a bijection $f_n\colon X_n\to\mathbb{N}$. (This step requires the Axiom of Countable Choice);
2. Select a bijection $g\colon\mathbb{N}\times\mathbb{N}\to\mathbb{N}$; there are several explicit examples of this. For example, the Cantor pairing function $g(p,q) = \frac{(p+q)(p+q+1)}{2}+q$.
3. Define $f\colon \bigcup\limits_{n\in\mathbb{N}}(X_n\times\{n\})\to \mathbb{N}$ by mapping $(x,n)$ to $g(f_n(x),n)$.

This defines a bijection between the disjoint union of the $X_n$ onto $\mathbb{N}$. To get a bijection in the case where the $X_n$ are not disjoint, note that $\bigcup\limits_{n\in\mathbb{N}} X_n$ embeds into the disjoint union (map $x$ in the union to $(x,m)$ where $m$ is the smallest $n\in\mathbb{N}$ such that $x\in X_n$), which is bijectable to $\mathbb{N}$; then use the Cantor-Bernstein Theorem applied to this embedding and to embedding that maps $\mathbb{N}$ to $X_1$ into the union to get a bijection.

However, if your set theory does not include the Axiom of Choice, then the answer may be that the union need not be bijectable with $\mathbb{N}$. In particular, it is consistent with ZF that the real numbers are a countable union of countable sets, and of course the real numbers are not bijectable with $\mathbb{N}$.

If you want to show that the countable union of countable subsets is countable, you can use Cantor-Schroeder-Bernstein (I don't think it uses AC --even in summer :) ), and set up injections between $\mathbb N$ and $\mathbb N \times \mathbb N $, and the other-way-around , by generalizing this:

take any two primes , say 2,3, and map : $(a,b)\rightarrow 2^a3^b$ ( you can see that, to generalize to a product of k-copies of $\mathbb N$, just take k different primes; if you want an actually countably-infinite product, this is maybe more delicate), and an injection in the opposite direction is given by , e.g., n->(n,0,0,...).

And, BTW, any choice of injections in CSBernstein allows to construct an actual bijection.

EDIT: I think it is not too hard to show the map (a,b)->$2^a3^b$ is an injection; if we had $2^a3^b=2^{a'}3^{b'}$, it would follow that $2^{a-a'}3^{b-b'}=1$; by simple divisibility arguments, each of the factors on the left-hand side would have to divide 1; it then follows that a-a'=0 and b-b'=0, i.e., a=a', b=b'.

EDIT#2 : Please see some of the caveats in the comments section about concluding that the union of countables is countable.