$\textbf{My solution} : $
We will use this theorem: If a set $X$ only has isolated points then $X$ is countable or equivalently if $X$ is uncountable then $X$ contains some point of accumulation in $X$ ($X \cap X'\neq \emptyset $)
Suppose that $X'$ is countable, then $Y=X-X'$ is uncountable (if it were we would have X is countable) then by the theorem we would have that there is some $a \in Y'\cap Y$.
We affirm that $a$ is point of accumulation of $X$.
Indeed, for any $\epsilon >0 : (a- \epsilon,a+ \epsilon) \cap Y-\{a\} \neq \emptyset$, then exists $x_1 \in (a- \epsilon,a+ \epsilon) \cap Y-\{a\}$. As $x_1$ is in $Y$ then $x_1$ is in $X$ so $x_1 \in (a- \epsilon,a+ \epsilon) \cap X-\{a\}$ then $a \in X'$ a contradiction because $a \in Y$.
I wanted to know if my solution is correct, something to improve or maybe some other. Thank you
Suppose $X$ is uncountable. Let $c(X)$ be the set of condensation points of $X$: $\{x \in X: \forall r>0: (x-r,x+r) \cap X \text{ uncountable}\}$. Let $B_n, n \in \mathbb{N}$ be a countable base for $\mathbb{R}$ (e.g. all rational intervals).
All condensation points are trivially in $X' \cap X$. For every point $p \in X\setminus c(X)$ there is some $n(p) \in \mathbb{N}$ such that the neighbourhood $B_{n(p)} \cap X$ of $p$ is at most countable. Then $X \setminus c(X)=\bigcup\{B_{n(p)}: p \in X\setminus c(X)\}$ is a countable union of countable open sets (of $X$) and so as $X$ is uncountable we have that $c(X)$ is closed in $X$ and uncountable. So in particular $X'$ is uncountable (and non-empty, which is all you needed to prove).