A space is completely normal if and only if every subspace is normal

Suppose $X$ is completely normal, let $S\subseteq X$, and let $A,B$ be disjoint closed subsets of $S$. By [F2] they are separated in $X$, so ambient disjoint open sets containing them trace to disjoint open sets of $S$. Thus $S$ is normal.

Conversely suppose every subspace of $X$ is normal, and let $A,B\subseteq X$ be separated. Put $Y=X\setminus((\overline A\setminus A)\cup(\overline B\setminus B))$; separation ensures that $A,B\subseteq Y$, and they are disjoint closed subsets of $Y$.

Normality of $Y$ gives disjoint open $U,V\subseteq Y$ containing $A,B$. Write $U=G\cap Y$ and $V=H\cap Y$ with $G,H$ open in $X$; then $G\cap H$ is contained in $(\overline A\setminus A)\cup(\overline B\setminus B)$.

The open sets $G\setminus\overline B$ and $H\setminus\overline A$ contain $A$ and $B$ respectively, and are disjoint: a point of their intersection would lie in $G\cap H$ but in neither of the two displayed closure differences.

Hence every separated pair in $X$ has disjoint open neighbourhoods, so $X$ is completely normal.