The Tietze extension theorem implies the Urysohn lemma

Let \(A,B\subset X\) be disjoint closed subsets of a normal space \(X\). The union \(A\cup B\) is closed. Define \[ f:A\cup B\to[-1,1] \] by \(f=0\) on \(A\) and \(f=1\) on \(B\). This is continuous because \(A\) and \(B\) are disjoint closed-and-open subsets of the subspace \(A\cup B\).

By the Tietze extension theorem, \(f\) extends to a continuous \[ F:X\to[-1,1]. \] Composing with the continuous truncation \[ t(s)=\min\{1,\max\{0,s\}\} \] gives \(u=t\circ F:X\to[0,1]\), with \(u(A)=0\) and \(u(B)=1\). This is exactly the Urysohn lemma.