Proof. Let $A,B\suf X$ be disjoint closed sets. By Theorem 3.13.10, there exists a partition of unity $\curl{f_i}_{i\in I}$ subordinate to $\curl{A^c,B^c}$. Let $I=I_{A}\sqcup I_{B}$ such that for each $i\in I_{A},\supp{f_i}\suf B^{c}$ and for each $i\in I_{B},\supp{f_i}\suf A^{c}$. Let $f=\sums{}{i\in I_A}f_{i}$ and $g=\sums{}{i\in I_B}f_{i}$ then $f|_{B}=0,g|_{A}=0$ hence $f|_{A}=1$. By continuity of $f$, $\curl{f\geq 2/3}\in \cali{N}_{X}(A)$ and $\curl{f\leq 1/3}\in\cali{N}_{X}(B)$ are disjoint as desired.$\square$