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. 

It’s enough to assume that $E$ has no limit points in $E$. Then for each $x\in E$ there is an $n(x)\in\Bbb N$ such that $(x-2^{-n(x)},x+2^{-n(x)})\cap E=\{x\}$; call this interval $U_x$. The rationals are dense in $\Bbb R$, so for each $x\in E$ we may choose a rational $q_x\in U_x$. Suppose that $E$ is uncountable. $\Bbb Q$ is countable, so there must be some $p\in\Bbb Q$ such that $E_0=\{x\in E:q_x=p\}$ is uncountable. And $\Bbb N$ is also countable, so there must be some $m\in\Bbb N$ such that $E_1=\{x\in E_0:n(x)=m\}$ is uncountable. But then $p\in U_x=(x-2^{-m},x+2^{-m})$ for each $x\in E_1$; equivalently, $|x-p|<2^{-m}$ for each $x\in E_1$. 
Suppose that $x,y\in E_1$ with $p\le x<y$; then $x\in E\cap[p,y]\subseteq E\cap U_y$, contradicting the choice of $U_y$. A similar contradiction arises if there are $x,y\in E_1$ such that $x<y\le p$. Thus, $|E_1\cap[p,\to)|\le 1$ and $|E_1\cap(\leftarrow,p]|\le 1$, so $|E|\le 2$, contradicting the uncountability of $E_1$ and showing that $E$ must in fact be countable.
