How many closed separable subsets does a Hausdorff space have?

Question

I have been having trouble trying to solve this one topology problem. It states "Let $X$ be a Hausdorff space whose cardinality is at most $c$, the cardinality of $\mathbb{R}$. Prove that $X$ has at most $c$ closed separable subsets." I have no idea how to go about proving this statement.

Answer 1

A closed subspace $A$ is separable iff it is of the form $A = \overline{C}$ where $C$ is an at most countable subset of $X$.

So $C \to \overline{C}$ is a surjection from $[X]^{\le \omega}$ onto $\{A \subseteq X: A \text{ closed and separable }\}$, where $[X]^{\le \omega}$ is the set of at most countable subsets of $X$. So the number of closed and separable subspaces is at most $|[X]^{\le \omega}|$.

Now standard set theory shows that if $|X| \le \mathfrak{c}$, then $|[X]^{\le \omega}| \le \mathfrak{c}^{\aleph_0} = \mathfrak{c}$. The last equality follows from $\mathfrak{c} = 2^{\aleph_0}$.

I don't see that why we need Hausdorffness for this. Maybe somebody else does?