I want to show that $E= \{ \frac{1}{n^2} : n \in \mathbb{N} \} \cup \{0\}$. I know from a theorem that $S= \{ \frac{1}{n} : n \in \mathbb{N} \} \cup \{ 0 \}$ is compact. Would the proof just be analogous then? Here's what I have:
Let $\mathcal{G}$ be an open cover of E. Since $0 \subset E \subset \cup_{G \in \mathcal{G}} G $, $\exists G_0 \in \mathcal{G}$ such that $0 \in \mathcal{G}$. Since $G_0$ is open, we have that 0 is an interior point of $G_0$. By definition of interior point, we then have that $N_r(0) \in G_0$.
I know that $\frac{1}{n^2} \in N_r(0)$ if $\frac{1}{n^2} < r \rightarrow n > \sqrt{\frac{1}{r}}$.
So define $N > \sqrt{\frac{1}{r}}$. If $n>N, n\in \mathbb{N}$, then $n\geq N+1 > \sqrt{\frac{1}{r}}$, and so $\frac{1}{n^2} < r$, which then implies that $\frac{1}{n^2} \in N_r (0) \subset G_0$. And now we've shown that $G_0$ covers 0 as well as $\{ \frac{1}{n^2} : n \in \mathbb{N} \}$.
Now what's left is proving finite subcovers:
For each $1 \leq n \leq N$, $\frac{1}{n^2} \in E \in \cup_{G \in \mathcal{G}} G$, and so $\exists G_n \in \mathcal{G}$ such that $\frac{1}{n^2} \in G_n$. Let $\mathcal{H} = \{ G_1, ..., G_n\}$, and so we have a finite subcover. Since $\mathcal{G}$ was arbitrary, E is compact.$