If $A,B\subseteq X$ is closed in $X$ and $A\cap B=\emptyset$, then $\exists f:X\to [0,1]$ such that $f$ is continuous, $f(A)=\{0\}$ and $f(B)=\{1\}$ $\implies$ $X$ is normal.
Munkres’ defined normal space as $T_1$ and $T_4$. Showing $X$ is $T_4$ is trivial. Since $f$ is continuous, $f^{-1}([0,x)), f^{-1}((y,1])\in \mathcal{T}_X$, where $0\lt x\lt y\lt 1$. By elementary set theory, $A\subseteq f^{-1}(f(A))=f^{-1}(\{0\}) \subseteq f^{-1}([0,x))$ and $B\subseteq f^{-1}(f(B))=f^{-1}(\{1\})\subseteq f^{-1}((y,1])$. Hence $\exists f^{-1}([0,x)), f^{-1}((y,1])\in \mathcal{T}_X$ such that $A\subseteq f^{-1}([0,x))$, $B\subseteq f^{-1}((y,1])$ and $ f^{-1}([0,x))\cap f^{-1}((y,1]) =f^{-1}([0,x) \cap (y,1])=f^{-1}(\emptyset)=\emptyset$. Now how to show $X$ is $T_1$?
Best Answer
Munkres does not claim that your converse is true. He only says that if all pairs of disjoint closed sets can be separated by a continuous function, then all pairs of disjoint closed sets can be separated by disjoint open sets (which is fairly trivial).
Unfortunately notation is not really standardized in the literature. $T_1$-spaces are certainly those whose one-point subsets are closed, but spaces which allow separation of disjoint closed sets by disjoint open sets are sometimes denoted as $T_4$ and sometimes as normal. Moreover, some authors (like Munkres) use "$T_4$" and "normal" as synonyms because they only define this concept in the realm of $T_1$-spaces.
Anyway, separation of disjoint closed sets by disjoint open sets (or by continuous functions) does not imply $T_1$. As an example take a space $X$ with the trivial topology and more than one point. Then $X$ does not have a non-trivial pair of disjoint closed sets, thus it trivially satisfies the "closed sets separation condition", but it not $T_1$.