$X \subset \mathbb{R}^n$ such that all points in $X$ are isolated $\implies$ $X^{'}$ enumerable 
Prove or disprove: $X \subset \mathbb{R}^n$ such that all points in $X$ are isolated $\implies$ $X^{'}$ enumerable

I need resolve it in order to solve another question, but even this seems simples i couldn't resolve.
My attempt(I hope this is true): Let $S$ be the set of convergent sequences in $X$.
Define:
$f \colon X^{'} \to S$, for each $x \in X^{'}$,  $f(x) \in S$ is a sequence that converges to $x$.
Is easy to see that $f$ is injective, so I want to find a bijection $\phi \colon f(X^{'}) \to X$ injective, using that $f(x)$ is a sequence of islated points in $X$. I didn't find $\phi$.
Any help is welcome. Thank you.
 A: Unfortunately, what you’re trying to prove is false. Let 
$$X=\left\{\left\langle\frac{2m+1}{2^n},\frac1{2^n}\right\rangle:n\in\Bbb N\text{ and }m\in\Bbb Z\right\}\;;$$
every point of $X$ is isolated in $X$, but $X'$ is the whole $x$-axis.
Added: For that matter, you could just as well use the simpler set
$$\left\{\left\langle\frac{m}{2^n},\frac1{2^n}\right\rangle:n\in\Bbb N\text{ and }m\in\Bbb Z\right\}\;,$$
for which the convergence to the $x$-axis may be a little easier to see. $X$ is a subset of it that has at most one point on each vertical line.
A: Here's another one, in $\mathbb R$.  
$X = $ the set of midpoints of the complementary intervals of the Cantor set.  So every point of $X$ is the midpoint of an interval with no other points of $X$; that is every point is isolated.  But $X'$ is the Cantor set, uncountable.
A: Hint: all point of $X$ are isolated i.e there exists open balls $B_x(x,r_x>0)$ with $B(x,r_x)\cap B(y,r_y)$ is empty if $x\neq y$. Take an element $q_x\in Q^n\cap B(x,r_x)$ an define $f:X\rightarrow Q^n f(x)=q_x$. Use the fact that $Q^n$ is numerable and $f$ is injective.
Remark.
A point $x$ is isolated if there is a ball $B(x,r_x)$ such that $B(x,r_x)\cap X=\{x\}$ this is equivalent here (in $R^n$) that you can suppose for every $x$ that $B(x,r_x)\cap B(y,r_y)$ is empty for some $r_y$ if $x\neq y$ otherwise there is $z_n\in B(x,1/n)\cap B(y_n,1/n), y_n\in X, y_n\neq x$ which implies that  $(y_n),$ converges towards $x$ since $d(x,y_n)\leq d(x,z_n)+d(z_n,y_n)\leq {1\over {2n}}$.
