I am by no means an expert, and I would have posted this as a comment. However, I don't have the reputation, so I will write an answer. I seem to be finding one small issue with Henno Brandsma's answer, and I think I also have a fix.
First, I should note that I am following Silverman's translation of Kolmogorov and Fomin which defines a limit point as: ``A point $x$ in a topological space $T$ is a limit point of a set $M \subset T$ if every open set containing $x$ also contains infinitely many points of $M$''.
Now I think there is a problem here:
``For every $x\in A \setminus A′$ ($A′=$ the set of limit points of $A$, that you call $D$) we can pick a basic open set $B(x)$ such that $B(x)\cap A=\{x\}$''. What if there only exists a neighbourhood $B(x)$, where $B(x) \cap A= C(x)$, such that $x \in C(x)$, and $C(x)$ is finite. With the provided definition of a limit point, this would be impossible for a metric space (shown in this question) or a $T_1$-space, but I could not find anything regarding a second countable space.
Here is my fix for the proof of showing that $A \setminus A'$ is countable: For every $x\in A \setminus A′$, we can pick a basic open set $B(x)$ such that $B(x)\cap A=C(x)$, such that $x \in C(x)$, and $C(x)$ is finite. Since all $C(x)$ are finite, there exists the finite integer $N=\max_{x\in A} |C(x)|$.
There exists a injection from $A\setminus A'$ to $\mathscr{B} \times N = \{(B(x),i) | x \in A, 1 \leq i \leq N \}$ by mapping each $x$ to $(B(x),i)$ for some $i \leq N$, as every $B(x)$ cannot contain more that $N$ members of $A$. As the set of bases, $\mathscr B$, is countable, so would be $\mathscr B \times N$. Therefore, $A\setminus A'$ should be countable as well.