I am asked to prove subject along with its converse (which I can do). I would appreciate any assistance based on my progress below in proving the separation conditions hold in the situation I have constructed. Thanks!
DEFINITIONS
$\textbf{completely normal}$: every subspace is normal
$\textbf{separated}$: $A$ and $B$ are separated if and only if $\overline{A}\cap B=\varnothing=A\cap\overline{B}$
PROVE: If for every pair of separated sets $A$ and $B$ in $X$ there exist disjoint open sets containing them, then $X$ is completely normal.
IDEAS:
Let $Y$ be a subspace of $X$ and let $A$ and $B$ be disjoint closed sets in $Y$. It must be shown there exist disjoint open sets $U$ and $V$ in $Y$ containing $A$ and $B$ (respectively). First show $A$ an $B$ are separated.
$\textit{This is where I need help...}$
idea 1: Since $A$ and $B$ are closed in $Y$, there exist closed sets $A_X$ and $B_X$ in $X$ such that $A_X\cap Y=A$ and $B_X\cap Y=B$. Since $A_X$ and $B_Y$ are closed in $X$, then $\overline{A_X}=A_X$ and $\overline{B_X}=B_X$.
idea 2: The sets $A$ and $B$ may not be closed in $X$, so let $\overline{A}$ and $\overline{B}$ be their closures in $X$. Then $\overline{A}\cap Y=A$ and $\overline{B}\cap Y=B$.
$\textit{This is where I need to prove $A$ and $B$ are separated...}$
Since $A$ and $B$ are separated, there exist disjoint open sets $U'$ and $V'$ in $X$ containing $A$ and $B$ respectively by hypothesis. Then $U=U'\cap Y$ and $V=V'\cap Y$ are disjoint open sets in $Y$ containing $A$ and $B$ respectively so $Y$ is normal and $X$ is completely normal since $Y$ was an arbitrary subset of $X$.
