TITLE: A first-countable, separable Hausdorff space has at most the continuum cardinality $c$
QUESTION [5 upvotes]: 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.

REPLY [10 votes]: Fix a countable dense subset 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).