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: 
Probably the simplest argument is that a singleton point set in $\mathbb{R}$ is closed, and finite unions of closed sets are closed. Therefore every finite set is closed, and the contrapositive of this says:

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> If set $E$ (in the real numbers with usual topology) is not closed, then $E$ is not a finite set.
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It is fairly easy to show a singleton point set $\{x\} \subset \mathbb{R}$ is closed. One argument would be to note that the complement $\mathbb{R} \setminus \{x\}$ is open, e.g. the union of two open intervals $(-\infty,x)\cup (x,+\infty)$.

Alternatively, if the finitely many points of a set $E$ were ordered:

$$ x\_1 \lt x\_2 \lt \ldots \lt x\_n $$

then we could argue $\mathbb{R}\setminus E = (-\infty,x\_1)\cup (x\_1,x\_2) \cup \ldots \cup (x\_n,+\infty)$ is open (as the union of open intervals). Then $E$ is again closed.
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.