For me at least, the first thing to do, especially for problems of this level, is to convince yourself that the theorem is true. One way to do that here would be to draw a picture. If you draw a Venn Diagram with a set $A$ and a set $B$ such that $A\subseteq B$, this theorem should make sense.
Once you have convinced yourself it is true and are kinda starting to see why, you can proceed with the proof. I should say, though, that sometimes the problems are too abstract to really convince yourself. For more advanced math, it can sometimes even be hard to see what is going on, at which point it might be prudent to think about whatever theorem you know that might apply, and blindly follow. But here this sort of thinking should be fine. Hopefully you know that this is an iff statement; that is, you need to prove that the first part implies the second part, and the second implies the first. To do that, you must assume that one holds and show the other must hold.
So here is one direction. Let's try and show that $A\subseteq B$ implies $A\setminus B=\emptyset$. As others have mentioned, let's assume $A\subseteq B$ and then show that the opposite cannot hold--that is, it cannot be that $A\setminus B\neq \emptyset$. So we assume that, and show there is a problem. If $A\setminus B\neq \emptyset$, by definition, there exists some element $x\in A$ such that $x\notin B$. But by definition, $A \subseteq B$ if and only if for all $x \in A$, $x \in B$. But we have shown there is such an element without this property, so it cannot be that $A\subseteq B$, which contradicts our original assumption that this held. Also note that this can also be formulated as a proof of the contrapositive instead of a proof by contradiction. Try to do the other direction in a similar way.