A finite union of nowhere dense sets is nowhere dense; a countable union need not be.Suggest a slogan
Let \(X\) be a topological space and let \(S_1,\ldots,S_k\subseteq X\) be nowhere dense. Then \(S_1\cup\cdots\cup S_k\) is nowhere dense.
By induction it suffices to treat \(k=2\). Since \(\overline{S_1\cup S_2}=\overline{S_1}\cup\overline{S_2}\), it suffices to show that a union of two closed sets \(F_1,F_2\) with empty interior has empty interior. Let \(U\subseteq F_1\cup F_2\) be open. Then \(U\setminus F_1\) is open and contained in \(F_2\), so \(U\setminus F_1=\emptyset\) and \(U\subseteq F_1\); hence \(U=\emptyset\).
The statement fails for countable unions: \(\QQ=\bigcup_{q\in\QQ}\theset{q}\) is a countable union of nowhere dense subsets of \(\RR\), but \(\QQ\) is dense in \(\RR\).