Following the idea of Theorem 27.7. If $A\subset X$ is closed, $U\subset X$ is non-empty and open and $U\not\subset A$ , then there is a non-empty open set $V\subset X$ such that $\overline{V}\subset U-A$ . Indeed, since $U-A\neq\emptyset$ , there is a point $x\in U-A$ , and there are disjoint open neighborhoods $W$ and $V$ about $A\cup(X-U)$ (closed) and $x$ , respectively (Lemma 26.4), so that $x\in V\subset\overline{V}\subset X-(A\cup(X-U))=(X-A)\cap U=U-A$ .

We take any non-empty open set $U_{0}$ and show that it has a point not in the union $\cup_{i\ge1}A_{i}$ . For $A_{i}$ , $i\ge1$ , $U_{i-1}$ is non-empty and $U_{i-1}\not\subset A_{i}$ ($A_{i}$ has no interior points), hence, there is a non-empty open set $U_{i}\subset X$ such that $\overline{U}_{i}\subset U_{i-1}-A_{i}$ . We have a nested sequence of non-empty closed sets $\overline{U}_{i}$ , and their intersection is nonempty as $X$ is compact, where any point in the intersection belongs to $\overline{U}_{1}\subset U_{0}$ but does not belong to $\cup_{i\ge1}A_{i}$ .