Every regular Lindelöf space is normal

Let $\mathcal{V}$ be the family of all open $V\subseteq X$ satisfying $\bar{V}\subseteq X\setminus B$. For every $a\in A$, [L1] supplies a member of $\mathcal{V}$ containing $a$, so $\mathcal{V}\cup \{X\setminus A\}$ is an open cover of $X$.

By Lindelöfness, an at most countable subfamily of $\mathcal{V}$ covers $A$. Interchanging $A$ and $B$ gives an at most countable family of open sets covering $B$, each with closure disjoint from $A$. List these families as $(U_n{)}_{n\ge 0}$ and $(V_n{)}_{n\ge 0}$, padding a finite family with empty sets. The countable-listing convention is part of [F1]. Only two subcovers and their listings are selected; no axiom of choice is needed.

Set $U={\bigcup }_{n\ge 0}(U_n\setminus {\bigcup }_{i\le n}\overline{V_i})$ and $W={\bigcup }_{n\ge 0}(V_n\setminus {\bigcup }_{i\le n}\overline{U_i})$. Each summand is open because it removes only finitely many closed sets from an open set. Every point of $A$ lies in some $U_n$ and in none of the $\overline{V_i}$, so $A\subseteq U$. Similarly $B\subseteq W$.

A point in the $n$th summand of $U$ and the $m$th summand of $W$ is impossible: if $m\le n$, the first summand excludes $\overline{V_m}$; if $n\le m$, the second excludes $\overline{U_n}$. Thus $U\cap W=\mathrm{\varnothing }$, and these open sets prove normality, including empty $A$ or $B$.