Prove that an open subset of a Baire space is a Baire space

A closed (in the subspace topology) subset $A$ of $U$ is generally not closed in $X$. Consider the subset $A = [0,1)$ of $U = (-1,1) \subset \mathbb{R} = X$.

But only a small modification is needed. Since the boundary of an open set has empty interior, it follows that for subsets of $U$, being nowhere dense in $U$ implies being nowhere dense in $X$. For, if $A \subset U$ is not nowhere dense in $X$, i.e. $V := \operatorname{int}_X(\operatorname{cl}_X(A)) \neq \varnothing$, then $W = V\cap U$ is a nonempty open set contained in $\operatorname{cl}_U(A) = U \cap \operatorname{cl}_X(A)$.

Thus, if $(A_n)$ is a sequence of relatively closed subsets of $U$ with empty interior, then $(\overline{A}_n)$ is a sequence of closed (in $X$) sets with empty interior, and $A_n = U \cap \overline{A}_n$ for all $n$.