Every uncountable subset of $\mathbb{R}$ has a limit point I am looking at this problem and I decided to attack it by proving the contrapositive.

If $E \subset \mathbb{R}$ has finitely limit points, then $E$ is countable.
My proof: Let $S=\{\text{limit points of E} \}$, then for any $s\in E, \ \exists \delta>0$ such that $B_{\delta}(s)$ contains a point of $E$. Take $\delta_1 >0, s_1 \in S$, then $B_{\delta_1}(s_1)$ contains a point of $E$ say $x_1$. We then take $\delta_2 >0, s_2, B_{\delta_2}(s_2)-{x_1}$ to contain $x_2$. Inductively we have $\{x_n\} \in (B_{\delta_{n}}(s_n)-\{x_1,...,x_n\})$. This gives us our bijective map:  $\phi: \mathbb{N} \to E \Rightarrow E$ is countable.
Question: I am unsure if what I am proving is infact the contrapositive. If this argument can be done directly I would like to know, if not and I am not doing the contrapositive, can someone quantify it for me? Am i going in the right direction or do I need to try something else.

 A: I think this is the proof by contrapositive that you were getting at.
Suppose that $E$ has no limit points at all. Pick an arbitrary point $x\in E$. Then $x$ cannot be a limit point, so there must be some $\delta>0$ such that the ball of radius $\delta$ around $x$ contains no other points of $E$: $$B_{\delta}(x)\cap E=\{x\}$$ Call this "point $1$". For the next point, take the closest element to $x$ and on its left; that is, choose the point $$\max \Big[E\cap (-\infty,x)\Big]$$ if it exists (that is important - if not, skip to the next step). Note that by the argument above, this supremum, should it exist, cannot equal $x$ and is therefore a new point in $E$.
Call this "point $2$". Now take the first point to the right of $x$ for "point $3$". Take the first point to the left of point $2$ for "point $4$". And so on, ad infinitum.
This gives a countable list of unique points; we must show that it exhausts the entire set $E$. Suppose not. Suppose there is some element $a<x$ which is never included in the list (picking $a$ on the negative side of $x$ is arbitrary, and the same argument would work for the second case). Then the element closest and to the right of $a$ in $E$ (which exists, by the no-limit-points argument at the beginning) is also not in the list; if it was, $a$ would have been in one of the next two spots. And same with that point (call it $a_1$); there is a closest $a_2>a_1\in E$ such that $a_2$ is not in the list. Repeating, we generate an infinite monotone-increasing sequence $\{a_i\}$ of elements in $E$ and not in the list, which is clearly bounded above by $x$. By the Monotone Convergence Theorem this sequence has a limit. But that means the sequence $\{a_i\}\subset E$ converges to a limit, and hence $E$ has a limit point, contradicting the assumption. Therefore our list exhausts $E$, and we have enumerated all its elements.
