I am redoing a demonstration of Set-Theoretic Topology and I can not prove the following:
Let $X$ be a Hausdorff and Separable topological space and $S$ a dense subset. Prove that if $p$, $q$ $\in X$ are distinct points, there exists $A \subset S$ such that p belongs to the closure of A and q does not belong.
If $U$ is an open set containing $p$ and $V$ an open set containing $q$ such that $U \cap V = \emptyset$, then consider $A = S \cap U$.
$q\notin \overline{A}$ as $V$ is a neighbourhood of $q$ disjoint from $A$, while supposing that $O$ is any open set containing $p$, so is $U \cap O$ and we know that $S$ must intersect the non-empty open set $U \cap O$ and so $O$ intersects $A$, showing that $p \in \overline{A}$.
If $X$ is separable, then we can take $S$ countable, and the map that sends $(p,q)$ to this subset shows that $|X|=|X|^2 \le |\mathscr{P}(S)| = \mathfrak{c}$, so we get an upper bound on the szie of $X$. Note that on a $T_1$ space you can get no such bound, as the cofinite topology on any size $X$ is separable and $T_1$. So Hausdorffness is essential.