Prove that the union of countably many countable sets is countable.
Let's start with a quick review of "countable". A set is countable if we can set up a 1-1 correspondence between the set and the natural numbers. As an example, let's take $\mathbb{Z}$, which consists of all the integers. Is $\mathbb Z$ countable?
It may seem uncountable if you pick a naive correspondence, say $1 \mapsto 1$, $2 \mapsto 2 ...$, which leaves all of the negative numbers unmapped. But if we organize the integers like this:
$$0$$ $$1, -1$$ $$2, -2$$ $$3, -3$$ $$...$$
We quickly see that there is a map that works. Map 1 to 0, 2 to 1, 3 to -1, 4 to 2, 5 to -2, etc. So given an element $x$ in $\mathbb Z$, we either have that $1 \mapsto x$ if $x=0$, $2x \mapsto x$ if $x > 0$, or $2|x|+1 \mapsto x$ if $x < 0$. So the integers are countable.
We proved this by finding a map between the integers and the natural numbers. So to show that the union of countably many sets is countable, we need to find a similar mapping. First, let's unpack "the union of countably many countable sets is countable":
"countable sets" pretty simple. If $S$ is in our set of sets, there's a 1-1 correspondence between elements of $S$ and $\mathbb N$.
"countably many countable sets" we have a 1-1 correspondence between $\mathbb N$ and the sets themselves. In other words, we can write the sets as $S_1$, $S_2$, $S_3$... Let's call the set of sets $\{S_n\}, n \in \mathbb N$.
"union of countably many countable sets is countable". There is a 1-1 mapping between the elements in $\mathbb N$ and the elements in $S_1 \cup S_2 \cup S_3 ...$
So how do we prove this? We need to find a correspondence, of course. Fortunately, there's a simple way to do this. Let $s_{nm}$ be the $mth$ element of $S_n$. We can do this because $S_n$ is by definition of the problem countable. We can write the elements of ALL the sets like this:
$$s_{11}, s_{12}, s_{13} ...$$ $$s_{21}, s_{22}, s_{23} ...$$ $$s_{31}, s_{32}, s_{33} ...$$ $$...$$
Now let $1 \mapsto s_{11}$, $2 \mapsto s_{12}$, $3 \mapsto s_{21}$, $4 \mapsto s_{13}$, etc. You might notice that if we cross out every element that we've mapped, we're crossing them out in diagonal lines. With $1$ we cross out the first diagonal, $2-3$ we cross out the second diagonal, $4-6$ the third diagonal, $7-10$ the fourth diagonal, etc. The $nth$ diagonal requires us to map $n$ elements to cross it out. Since we never "run out" of elements in $\mathbb N$, eventually given any diagonal we'll create a map to every element in it. Since obviously every element in $S_1 \cup S_2 \cup S_3 ...$ is in one of the diagonals, we've created a 1-1 map between $\mathbb N$ and the set of sets.
Let's extend this one step further. What if we made $s_{11} = 1/1$, $s_{12} = 1/2$, $s_{21} = 2/1$, etc? Then $S_1 \cup S_2 \cup S_3 ... = \mathbb Q^+$! This is how you prove that the rationals are countable. Well, the positive rationals anyway. Can you extend these proofs to show that the rationals are countable?
@Hovercouch's answer is correct, but the presentation hides a really rather important point that you ought probably to know about. Here it is:
The argument depends on accepting (a weak version of) the Axiom of Choice!
Why so?
You are only given that each $S_i$ is countable. You aren't given up front a way of counting any particular $S_i$, so you need to choose a surjective function $f_i\colon \mathbb{N} \to S_i$ to do the counting (in @Hovercouch's notation, $f_m(n) = s_{mn}$). And, crucially, you need to choose such an $f_i$ countably many times (a choice for each $i$).
That's an infinite sequence of choices to make: and it's a version of the highly non-trivial Axiom of Choice that says, yep, it's legitimate to pretend we can do that.
Lema 1. The union of two countable sets is countable.
Proof. Let $A = \{a_n : n \in \mathbf N\}$ and $B = \{b_n : n \in \mathbf N\}$. Then we can define the sequence $(c_n)_{n=0}^\infty$ by $$c_{2k} = a_k \quad\text{and}\quad c_{2k+1} = b_k$$ for every $k \in \mathbf N$. Now $A \cup B = \{c_n : n \in \mathbf N\}$ and since it is a infinite set then it is countable.
By Lemma 1 you can prove your proposition by induction on the number of sets of the family
Corollary. The union of a finite family of countable sets is a countable set.
To prove for a infinite family you need the Axiom of choice.