Showing that a space is completely normal iff every subspace is normal

Solution 1:

$\newcommand{\cl}{\operatorname{cl}}$Your $\Rightarrow$ argument is fine, save that it would be clearer if you said explicitly that $\overline A$ denotes closure in $X$. (This is why I prefer the notation $\cl A$, since it is so easily modified to show the space in which the closure is being taken: $\cl_XA$.)

You $\Leftarrow$ argument isn’t right: $X$ is a subspace containing $\cl A\cup\cl B$, so you’re not actually using the hereditary normality of $X$ at all. Let $Y=X\setminus(\cl_XA\cap\cl_XB)$; $\cl_YA$ and $\cl_YB$ are disjoint closed subsets of $Y$, and $Y$ is both normal and an open subset of $X$, so ... ?