Subset of separable metric space can have at most a countable amount of isolated points

Let $(X,d)$ be a separable metric space. Prove that every subset $Y \subset X$ can have at most a countable amount of isolated points.

Attempt at proof: Let $Y$ be an arbitrary (non-empty) subset of $X$. Since $(X,d)$ is separable, there exists a countable dense subset $A \subset X$ such that $\overline{A} = X$. Since $A$ is dense in $X$, we have that $A \cap Y \neq \emptyset.$ For every isolated point $x \in A \cap Y$ there exists then a $\delta_x > 0$ such that $B(x,\delta_x) \cap Y = \left\{x\right\}$.

Now I don't know how to prove the isolated points are countable. Any help?