Proof. Since $K$ is compact, assume without loss of generality that $\seqi{U}= \seqf{U_j}$.

For every $x \in K$, there exists $1 \le j \le n$ and $N_{x} \in \cn(x)$ compact such that $x \in N_{x} \subset U_{j}$. By compactness of $K$, there exists $\seqf[m]{x_j}\subset K$ such that $K = \bigcup_{j = 1}^{m} N_{x_j}$.

For each $1 \le j \le n$, let

then $F_{j} \subset U_{j}$ is compact, and $\bigcup_{j = 1}^{n} F_{j} \supset K$.

By Urysohn’s lemma, there exists $\seqf{f_j}\subset C_{c}(X; [0, 1])$ such that for each $1 \le j \le n$, $f_{j}|_{F_j}= 1$, and $\supp{f_j}\subset U_{j}$.

By Urysohn’s lemma again, there exists $f_{j + 1}\in C(X; [0, 1])$ such that $f_{j+1}|_{K}= 0$ and $\bracs{f_{j+1} = 0}\subset \bigcup_{j = 1}^{n} \supp{f_j}$. Let $F = \sum_{j = 1}^{n+1}f_{j}$, then $F(x) > 0$ for all $x \in X$. For each $1 \le j \le n$, let $g_{j} = f_{j}/F$, then $g_{j} \in C_{c}(X; [0, 1])$ with $\supp{g_j}\subset U_{j}$. In addition, since $f_{j+1}|_{K} = 0$,

Therefore $\seqf{g_j}$ is the desired partition of unity.$\square$