A metric space is called separable if it contains a countable dense subset.
I have no idea how to go about proving this. What sort of things should I understand to do this. What would a sample proof look like?
Well, you're looking for something that's countable and dense. The natural thing to look at would be $\mathbb Q$. In this case, look at $\mathbb Q ^K$. It's countable as it's a finite crossproduct of countable sets. To show it's dense, you have to show that every point in $\mathbb R ^k$ is a limit point of $\mathbb Q^k$. For that, use a sequence approach (Decimal approximations is the natural one)
Since this is also a question with "general-topology" tag:
Theorem: Let X and Y be separable topological spaces. Then $X \times Y$ is separable.
Proof: There are two countable sets $A \subseteq X $ and $B \subseteq Y$ such that $\overline A=X$ and $\overline B =Y$. Since A and B are countable $A \times B$ is countable too.
For an arbitrary $(x,y) \in X \times Y $ holds: For every open neighborhoods $U,V$ of $x$ and $y$ the intersections $U\cap A$ and $V\cap B$ are not empty, hence the intersection $U \times V \cap A \times B$ is not empty too.
From this it follows that the intersection of $A \times B$ with every open neighborhood containing $(x,y)$ is non-empty.