This is Munkres Chapter 4, Section 30, Problem Number 3.
Let $X$ have a countable basis; let $A$ be an uncountable subset of $X$. Show that uncountably many points of $A$ are limit points of $A$.
My solution:
I will show that there are only at most countable points of $A$ that are not a limit point of $A$. Since $X$ is second-countable, its subspace $A$ is second-countable. For a point $x \in A$ which is not a limit point of $A$, there exists a neighborhood $U$ of $x$ such that $A \cap U = \{ x \}$. Therefore $\{ x \}$ is open in $A$, and the countable basis $\mathcal{B}$ of $A$ contains the set $\{ x \}$. For every $x \in A$ which is not the limit point of $A$, there exists a unique one-point set in $\mathcal{B}$. Since $\mathcal{B}$ is countable, there are only at most countable number of points of $A$ which is not the point of $A$, and therefore there are uncountably many limit points in uncountable set $A$.
Does that make sense without any critical gaps?
The short summary of your proof: the set $B$ of non-limit points of $A$ would be uncountable and discrete as a subspace and this cannot be as it is second countable like $X$ is.
In short the idea of the proof is fine, but it could do with a better write-up. That an uncountable and discrete subspace is not second countable is what you're trying to say but not saying directly (the proof indeed being that all singletons must be in any base for the discrete subspace). Try to more direct. $A$ is the disjoint union of $A \cap A'$, its limit points, and the set I called $B$. If the former is countable the latter must be uncountable etc.