The Stacks project

Proof. It suffices to prove the lemma for two nowhere dense sets as the result in general will follow by induction. Let $A,B \subset X$ be nowhere dense subsets. We have $\overline{A \cup B} = \overline{A} \cup \overline{B}$. Hence, if $U \subset \overline{A \cup B}$ is an open subset of $X$, then $U \setminus U \cap \overline{B}$ is an open subset of $U$ and of $X$ and contained in $\overline{A}$ and hence empty. Similarly $U \setminus U \cap \overline{A}$ is empty. Thus $U = \emptyset $ as desired. $\square$