Prove if a discrete subset $E$ of $\mathbb{R}^n$ is compact, then it is finite. 
Let $E\subseteq \mathbb{R}^n$ be a compact set and suppose that for
each $x\in E$, there is an $r_x \gt 0$ such that $E \cap B_{r_x}(x) = \{x\}.$
Prove that $E$ is a finite set.

I am a bit confused as to what "finite" means here. Do I have to prove that $E$ is closed and bounded? This is giving me memories of the Heine-Borel Theorem. Am I on the right track?
 A: $E$ is covered by the balls $B_{r_x}(x)$. By compactness there is a finite subcover, say $E \subseteq \bigcup_{i=1}^{m} B_{r_{x_i}}(x_i)$. So $E \subseteq \{x_1,x_2,...,x_m\}$ proving that $E$ has finite number of elements.
A: Let $E\subseteq \mathbb{R}^n$ be a compact set and suppose that for each $x\in E$, there is an $r_x\gt0$ such that $E \cap B_{r_x}(x)$ = {x}.
Then all the points $x \in E$ are isolated points. If E is infinite set, E would have infinite isolated points, where that can not ne the case on the metric space $\mathbb{R}^n$. Because $\mathbb{R}^n$ is a separable compact metric space.
Take this problem as inlightness, and you can generlize it for the euclidian metric space:
Prove that every subset of $\mathbb{R}$ has countably many isolated points.
Let $A \subset \mathbb{R}$ and let $S$ the set of isolated points in $A$. Let $ x \in S$, so that there is an $ \epsilon >0 $ such that $(x-\epsilon, x+\epsilon) \cap A-\{x\}=\phi$. Now, let $a,b$ any two rational numbers such that $(x-\epsilon < a < x < b< x+\epsilon )$ or $x \in (a,b) \subset (x-\epsilon,x+\epsilon)$.
Define $\Phi: S \rightarrow \mathbb{Q}^2$ as $\Phi(x) = (a,b)$. Since there is  a correspondence between  the set $S$ and $\mathbb{Q}^2$. Thus $S$ is a countable set.
