So, what it says is that a completely regular space is "normal" on pairs of closed sets at least one of which is compact. Each point $x$ of the compact set $A$ can be separated from the closed set $B$ by a continuous function $f_x$ (here $f_x(x)=0$ and $f_x(B)=1$ ). Then for some fixed $0<\epsilon<1$ , $U_x=f_x^{-1}([0;\epsilon))$ defines an open neighborhood of $x$ that does not intersect $B$ . $\{U_x\}$ covers $A$ . We can find a finite subcover $\{U_{x_k}\}$ , $k=1 . . . n$ , and take the corresponding finite product of continuous functions $f=f_{x_1}\cdot . . .\cdot f_{x_n}$ . Now, for each $x\in A$ : $x\in U_{x_k}$ for some $k$ , and $f_{x_k}(x)<\epsilon$ , therefore, $f(x)\in[0;\epsilon)$ . For every $x\in B$ we have $f(x)=1$ . All we need to do now is to transform $f$ : find continuous $h:[0,1]\rightarrow[0;1]$ such that for $g=h\circ f$ : $g(x)=0$ for $x\in A$ and $g(x)=f(x)=1$ for $x\in B$ . This will separate $A$ and $B$ . Indeed, the finite product of continuous functions is continuous and the composition of continuous functions is continuous. For example, $h(z)=(\max\{z,\epsilon\}-\epsilon)/(1-\epsilon)$ will do.