Every regular space with a countable basis 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 $T_3$), $\forall a\in A$, $\exists W_a \in \mathcal{N}_a$ such that $\overline{W_a} \cap B=\emptyset$. Since $a\in W_a \in \mathcal{T}_X$, we have $\exists U_a \in \mathcal{B}$ such that $a\in U_a\subseteq W_a$. By Exercise 6, Section 17 of Munkres’ Topology, $\overline{U_a}\subseteq \overline{W_a}$. So $\overline{U_a} \cap B =\emptyset$, $\forall a\in A$. So $\{U_a| a\in A\}\subseteq \mathcal{B}$ is an open cover of $A$. By subset of countable set is countable, $\{U_a| a\in A\}$ is countable. So $\{U_a| a\in A\} =\{U_n|n\in \Bbb{N}\}$. Similarly construct $\{V_n|n\in \Bbb{N}\}$ open cover of $B$ 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 better version of Munkres proof?
First paragraph of Munkres’ proof is bit convoluted, IMO. Simply take $U=X-B$. We have actually shown, if $X$ is $T_3$ and 2 countable, then $X$ is $T_4$, don’t we?
