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. 

Note: The union of countably many countable sets is countable.
Assume $E$ is not countable. Then one of the sets $[n,n+1]\cap E$ with $n\in \mathbb Z$ is uncountable.
Starting with $a_0=n$, $b_0=n+1$ we find a sequence of nested intervals $[a_k, b_k]$ such that $[a_k,b_k]\cap E$ is uncountable. In fact we can simply bisect an interval at each step and note that at least one of the halves must have iuncountably many points in common with $E$.
In other words, we let $a_{k+1}=a_k$, $b_{k+1}=\frac{a_k+b_k}2$ if $[a_k, \frac{a_k+b_k}2]$ is uncountable and let  $a_{k+1}=\frac{a_k+b_k}2$, $b_{k+1}=b_k$ otherwise (and observe that then $[a_{k+1},b_{k+1}]\cap E$ is uncountable as well.
 The nested intervals contain a point $c\in \mathbb R$. This $c$ has some (in fact uncountably many) points of $E$ in every $\epsilon$-neighbourhood. Indeed $[a_k,b_k]$ is contained in the $\epsilon$-neighbourhood as soon as $2^{-k}<\epsilon$.

Or:
If $E\cap [0,\infty)$ and $E\cap(-\infty,0]$ are both countable, then so is $E$.
Hence assume wlog that $E\cap [0,\infty)$ is uncountable.
Let $a=\inf\{x\in \mathbb R\colon [0,x]\cap E\mathrm{\ is\ uncountable}\}$ 
If $a=\infty$ then $E\cap [0,\infty)=\bigcup_n E\cap[0,n)$ is the union of countable sets, hence countable.
If on the other hand $a$ is finite, then $[a,a+\epsilon)\cap E$ is uncountable for any $\epsilon>0$, hence $a$ is a limit point.
