Need a hint: show that a subset $E \subset \mathbb{R}$ with no limit points is at most countable. I'm stuck on the following real-analysis problem and could use a hint:
Consider $\mathbb{R}$ with the standard metric. Let $E \subset \mathbb{R}$ be a subset which has no limit points. Show that $E$ is at most countable. 
I'm primarily confused about how to go about showing that this set $E$ is at most countable (i.e. finite or countable).
What I can show: since $E$ has no limit points, I can show that for every $x \in E$, there is a neighborhood $N_{r_x}(x)$ where $r_x > 0$ that does not contain any other point $y \in E$ where $y \neq x$. This suffices to show that every point within x is an isolated point. 

proof by contradiction.
suppose E is uncountable. consider the set E intersection with [n,n+1] for integers n.
then there exist at least one n such that the above type of set is uncountable. clearly that set is bounded in real nos. then by Bolzano Weierstrass's theorem the above set has a limit point in R. 
a contradiction to the hypothesis.
