Question: This question was from the previous midterm and I have no idea where to start. I was thinking of using the theorem that a closed set contains all of its limit points or if a set is closed then its compliment is open.

Answer: 
If $E$ is not closed, there is a limit point of $E$ which is not contained in $E$. Most importantly, there is a limit point of $E$. Finite sets have no limit points.
Welp, the easy way is that every singleton is closed. (Although it wouldn't hurt to argue why that is.) So ever finite union of singletons is closed. Every finite set is a finite union of singletons and thus closed. So if a set isn't closed it can't be finite.

But if you didnt think of that, as I didn't. The direct way is:

If $E$ is not closed then there is limit point $p$ of $E$ with $p \not \in E$. Let $\epsilon > 0$. We will construct a sequence of $\epsilon\_i$ and $q\_i \in E$ so that $\epsilon\_0 = \epsilon$. As $p$ is a limit point there is a $q\_0 \in N(\epsilon\_0, p)\cap E$. We will define $\epsilon\_1 = |p-q\_0|$ so $\epsilon\_0 > \epsilon\_1 > 0$. Inductively we will select $q\_i \in N(\epsilon\_i, p)\cap E$ and inductively we will set $\epsilon\_{i+1} = |p-q\_i|$. Thus by induction $\epsilon\_0 > \epsilon\_2 > .... > 0$ and there are infinite $q\_i$. So $\{q\_i\}\subset E$. So $E$ is infinite.