A first-countable, separable Hausdorff space has at most the continuum cardinality $c$

Question

Show that if $X$ is first-countable, separable Hausdorff topological space, then $X$ has at most the continuum cardinality.

I do not know how to start.

Answer 1

Fix a countable dense subset $D$ of $X$.

Because $X$ is first countable and every $x$ is in the closure of $D$, we can pick a sequence $(d_n(x))_n$ from $D$ that converges to $x$.

There are at most $|D|^\omega = \mathfrak{c}$ many such sequences. And the map sending $x$ to its sequence is 1-1 as $X$ is Hausdorff (sequences have unique limits).