# Kuratowski-Ulam Theorem, nowhere dense set in product space

Let $$E$$ be a nowhere dense subset of $$\mathbb{R}\times \mathbb{R}$$. For $$x\in \mathbb{R}$$, define $$E_x=\{ y\in\mathbb{R}\mid (x,y)\in E\}.$$ Let $$D$$ denote the set of $$x$$ for which $$E_x$$ is NOT nowhere dense in $$\mathbb{R}$$. By the Kuratowski-Ulam Theorem, we know that $$D$$ is of first cateogory in $$\mathbb{R}$$. My question is: can we strengthen this conclusion by saying that $$D$$ is nowhere dense in $$\mathbb{R}$$? If not, is there a counterexample?

List the rational numbers as $$\{q_n : n = 1, 2, \cdots \}$$ and for each $$n$$, let $$X_n = \{q_n\} \times [n, n+1]$$. Finally, let $$X = \cup_{n = 1}^\infty X_n$$. Then $$X$$ is nowhere dense in $$\mathbb R\times\mathbb R$$ and for $$x$$ a rational number, $$E_x$$ is a non-degenerate closed interval, so the set $$D$$ is the set of rational numbers which, of course, fails to be nowhere dense in $$\mathbb R$$.