A question on the regular space Here is an exercise:

Show that if $A$ and $B$ are disjoint closed subsets of a regualr space $X$ which both have the Lindelof property, then there exist open sets $U, V \subset X$ such that $A \subset U$, $B \subset V$ and $U \cap V=\emptyset$.

Thanks for any help.
 A: $\newcommand{\cl}{\operatorname{cl}}$The proof is just like the proof that a regular Lindelöf space is normal. Let $X$ be regular, and let $A$ and $B$ be disjoint closed subsets of $X$. For each $x\in A$ let $G_x$ be an open nbhd of $x$ such that $\cl G_x\cap B=\varnothing$, and for each $x\in B$ let $H_x$ be an open nbhd of $x$ such that $\cl H_x\cap A=\varnothing$. $A$ and $B$ are Lindelöf, so there are countable families $$\mathscr{G}=\{G_n:n\in\omega\}\subseteq\{G_x:x\in A\}$$ and $$\mathscr{H}=\{H_n:n\in\omega\}\subseteq\{H_x:x\in B\}$$ such that $A\subseteq\bigcup\mathscr{G}$ and $B\subseteq\bigcup\mathscr{H}$.
Let $U_0=G_0$ and $V_0=H_0\setminus\cl G_0$. Given $U_n$ and $V_n$ for some $n\in\omega$, let $U_{n+1}=G_{n+1}\setminus\cl\bigcup_{k\le n}V_k$ and $V_{n+1}=H_{n+1}\setminus\cl\bigcup_{k\le n+1}U_k$. 
Let $U=\bigcup_{n\in\omega}U_n$ and $V=\bigcup_{n\in\omega}V_n$; certainly $U$ and $V$ are open. If $x\in A$, let $n$ be minimal such that $x\in G_n$; then $x\in U_n$, since $x\notin\cl V_k$ for any $k\in\omega$, so $A\subseteq U$. A similar argument shows that $B\subseteq V$. Finally, suppose that $x\in U\cap V$. Then $x\in V_n$ for some $n\in\omega$, so $x\notin U_k$ for $k\le n$, and therefore $x\in U_k$ for some $k>n$. But then the construction of $U_k$ ensures that $x\notin V_n$, which is absurd. Thus, $U\cap V=\varnothing$.
