A space is normal iff every pair of disjoint closed subsets have disjoint closed neighbourhoods. 
A space is normal iff every pair of disjoint closed subsets have disjoint closed neighbourhoods.

Given space $X$ and two disjoint closed subsets $A$ and $B$.
I have shown necessity: If X is normal then by Urysohn's lemma there exists continuous map $f:X \to [0,1]$ such that $f(A)=0$ and $f(B)=1$, then $f^{-1}(0)$ and $f^{-1}(1)$ are two disjoint closed neighbourhoods of A and B.
But how to show the sufficiency?

If $A$ and $B$ have disjoint closed neighborhoods $U$ and $V$, then by definition of neighborhood we know that $A\subseteq \mathrm{int}(U)$ and $B\subseteq\mathrm{int}(V)$. Now, the interior of $U$ is an open neighborhood of $A$, the interior of $V$ is an open neighborhood of $B$, and so $A$ and $B$ have disjoint open neighborhoods.
Thus, if a space satisfies your requirement, then it is normal.
