Here's my attempt at a solution and I'm wondering if it's correct.
Let $X$ have a countable basis with $A \subset X$ an uncountable set. Show $A$ has uncountably many limit points.
Let $A'$ be the set of all limit points of $A$. Assume $A'$ is countable.
Then $A-A'$ is uncountable. Let $x\in A-A'$, then $x$ is not a limit point. Since $x$ is not a limit point by definition there exists some $U_x$ open s.t. $U_x \cap A = \{x\}$. If we have some $x' \neq x$ where $x' \in A-A'$ then there is some $U_{x'}$ where its intersection with $A$ is similarly $\{x'\}$. Therefore $U_x$ is distinct from $U_{x'}$. Since $A-A'$ is uncountable there are uncountably many distinct neighborhoods in $X$. Therefore $X$ cannot have a countable basis. This is a contradiction so there must be uncountably many limit points.