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?
 A: There’s a flaw in your argument for necessity: $f^{-1}[\{0\}]$ and $f^{-1}[\{1\}]$ may not be nbhds of $A$ and $B$, respectively. Consider, for example, $X=[0,1]$, $A=\{0\}$, $B=\{1\}$, and $f:X\to[0,1]:x\mapsto x$: $f^{-1}[\{0\}]=\{0\}$, which is not a nbhd of $\{0\}$, and $f^{-1}[\{1\}]=\{1\}$, which is not a nbhd of $\{1\}$.
You can fix it by using $f^{-1}\left[\left[0,\frac13\right]\right]$ and $f^{-1}\left[\left[\frac23,1\right]\right]$ instead: these really are closed nbhds of $A$ and $B$, respectively, since they contain the open nbhds $f^{-1}\left[\left[0,\frac13\right)\right]$ and $f^{-1}\left[\left(\frac23,1\right]\right]$.
Arturo has already taken care of the other half of the argument.
A: 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.
