 Solution

Let $A,B$ be disjoint closed subsets of the metric space $(X,d)$; the cases where one is empty are immediate. The distance functionsare continuous, because each is 1-Lipschitz. Their sum is strictly positive: if both distances vanished, closedness would put $x$ in $A\cap B$. Thereforeis continuous, equals zero on $A$, and equals one on $B$. The setsare disjoint open neighbourhoods of $A$ and $B$. Thus every metric space is a normal topological space; this is the normality of every metric space.