 Solution

The Urysohn lemma states that if $A$ and $B$ are disjoint closed subsets of a normal space $X$, there is a continuous $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

The Tietze extension theorem states that if $A$ is a closed subset of a normal space $X$, then every continuous $f:A\to \mathbb{R}$ has a continuous extension $F:X\to \mathbb{R}$. In the bounded form, if $f(A)\subseteq [a,b]$, the extension can be chosen with $F(X)\subseteq [a,b]$.