If $X$ is a completely normal topological space, then for every pair of separated sets $A,B$ (i.e., $\overline{A} \cap B = \emptyset = A \cap \overline{B}$), there exist disjoint open sets $U$ and $V$ such that $A \subseteq U$ and $B \subseteq V$ [hint: consider $X-(\overline{A} \cap \overline{B})]$
If I'm correctly understanding what the hint is 'hinting' at, I need to show that both $A$ and $B$ are closed in $X-(\overline{A} \cap \overline{B}) = (X - \overline{A}) \cup (X - \overline{B})$, since this would imply there exist disjoint open sets in $(X - \overline{A}) \cup (X - \overline{B})$ that separate $A$ and $B$; but since $(X - \overline{A}) \cup (X - \overline{B})$ is open, these disjoint open sets would also be open in $X$.
I tried showing that $\overline{A} \cap [(X - \overline{A}) \cup (X - \overline{B})] = A$, but this isn't necessarily true (take $A=(0,1)$ and $B=(1,2)$ in $\Bbb{R}$), although the LHS does contain $A$. How do I show that $A$ and $B$ are closed in $(X - \overline{A}) \cup (X - \overline{B})$?
All you need is that $A’=\overline{A}\cap Y$ and $B’=\overline{B}\cap Y$ are closed in $Y=X-(\overline{A}\cap \overline{B})$ (which is clear) and also disjoint, and that $A\subseteq A’$. Disjointness follows from the separatedness. Then apply normality $Y$ to these new sets.