Show that every regular Lindelöf space is normal.
My attempt: Let $A$,$B$ be closed in $X$ such that $A \cap B=\emptyset$. By chapter 7 theorem 2.2 of Dugundji topology (equivalent definition of $T_3$), $\forall a\in A$, $\exists U_a \in \mathcal{N}_a$ such that $\overline{U_a} \cap B=\emptyset$. By Exercise 9, Section 30 of Munkres’ Topology, $A$ and $B$ are lindelof space. $U=\{ U_a| a\in A\}$ is an open cover of $A$. So $\exists \{U_n| n\in \Bbb{N}\}$ countable subcover of $U$. Similarly, $\exists \{V_n|n\in \Bbb{N}\}$ countable subcover of $V=\{V_b|b\in B\}$ such that $\overline{V_n}\cap A=\emptyset$, $\forall n\in \Bbb{N}$. Define $U_n^{\prime} =U_n -\bigcup_{i=1}^{n}\overline{V_i}$ and $V_n^{\prime} =V_n -\bigcup_{i=1}^{n}\overline{U_i}$, for each $n \in \Bbb{N}$. By exercise 4 section 17, $U_n^{\prime}, V_n^{\prime} \in \mathcal{T}_X$, $\forall n\in \Bbb{N}$. Let $U^{\prime}= \bigcup_{n\in \Bbb{N}} U_n^{\prime}$ and $V^{\prime}= \bigcup_{n\in \Bbb{N}} V_n^{\prime}$. So $U^{\prime}, V^\prime \in \mathcal{T}_X$. We claim $A \subseteq \bigcup_{n\in \Bbb{N}} U_n^{\prime}$. Let $x\in A$. Since $A\subseteq \bigcup_{n\in \Bbb{N}} U_n$, $\exists m\in \Bbb{N}$ such that $x\in U_m$. Since $\overline{V_n}\cap A=\emptyset$, we have $x\notin \overline{V_n}$, $\forall n\in \Bbb{N}$. So $x\in U_m^{\prime} \subseteq \bigcup_{n\in \Bbb{N}} U_n^{\prime}$. similarly, $B\subseteq \bigcup_{n\in \Bbb{N}} V_n^{\prime}$. $U^\prime \cap V^\prime =\emptyset$. Assume towards contradiction, $\exists z\in U^\prime \cap V^\prime$. Then $z\in U_p^\prime \cap V_q^\prime$, for some $p,q\in \Bbb{N}$. WLOG, assume $p\lt q$. $z\in V_q^\prime =V_q -\bigcup_{i=1}^q \overline{U_i}$. So $z\notin \bigcup_{i=1}^q \overline{U_i}$. $z\notin \overline{U_p}$. Since $U_p \subseteq \overline{U_p}$, we have $z\notin U_p$. Which contradicts our initial assumption of $z\in U_p$. Thus $U^\prime \cap V^\prime =\emptyset$. Hence $\exists U^\prime, V^\prime \in \mathcal{T}_X$ such that $A\subseteq U^\prime$, $B\subseteq V^\prime$ and $U^\prime \cap V^\prime =\emptyset$. Is this proof correct?
Note: Here is slight variation of above proof. Claim: $U_n^{\prime} =U_n -\bigcup_{i=1}^{n}\overline{V_i}=U_n \cap (\bigcap _{i=1}^n X-\overline{V_i})$ and $V_n^{\prime} =V_n -\bigcup_{i=1}^{n}\overline{U_i}= V_n \cap (\bigcap _{i=1}^n X-\overline{U_i})$. Proof: Let $x\in U_n -\bigcup_{i=1}^{n}\overline{V_i}$. Then $x\in U_n$ and $x\notin \overline{V_i}$, $\forall i\in J_n$. So $x\in X-\overline{V_i}$, $\forall i\in J_n$. Thus $x\in U_n \cap (\bigcap_{i=1}^n X-\overline{V_i})$. Conversely, let $x\in U_n \cap (\bigcap_{i=1}^n X-\overline{V_i})$. Then $x\in U_n$ and $x\in X-\overline{V_i}$, $\forall i\in J_n$. So $x\notin \overline{V_i}$, $\forall i\in J_n$. Thus $x\in U_n - \bigcup_{i=1}^n \overline{V_i}$. Hence $U_n -\bigcup_{i=1}^{n}\overline{V_i}=U_n \cap (\bigcap _{i=1}^n X-\overline{V_i})$. Proof is similar for $V_n -\bigcup_{i=1}^{n}\overline{U_i}= V_n \cap (\bigcap _{i=1}^n X-\overline{U_i})$.
Application: Theorem 32.1 of Munkres’ Topology.
