TITLE: Prove that an open subset of a Baire space is a Baire space
QUESTION [6 upvotes]: Problem:
Prove that if $X$ is a Baire space and $U\subset X$ is an open set, then the subspace $U$ is a Baire space. 
My proof:
Let $ \{A_n \} $ be any countable collection of closed sets of $ U $, each of which has empty interior. Since $ U $ is open, all the sets $ A_n $ is closed in $ X $ as well. Since $ X $ is a Baire space, $ \bigcup A_n $ is has empty interior in $ X $. Since $ A_n\subset U $ for each $ n $, $ \bigcup A_n\subset U $ as well, and hence $ A_n $ has empty interior in $ U $. It follows that $ U $ is a Baire space. 
Question: Is my proof correct? I am claiming that a set closed in an open set is closed in $X$ as well. I that true?

REPLY [2 votes]: Here is a possibly easier proof in terms of the alternative definition for Baire spaces: a space in which the intersection of any countable family of open dense sets is dense.
Suppose we are given $O_n$ open in $U$ and dense in $U$.  The set $G_n=O_n\cup(X\setminus\overline{U})$ is open in $X$ and dense in $X$, since its closure contains $X\setminus\overline U$ and contains $\overline{O_n}\supseteq\overline{U}$.  Then $\bigcap_n G_n=\bigcap_n O_n\cup(X\setminus\overline U)$ is dense in $X$ by the Baire property of $X$.  So any nonempty open subset of $U$ must meet $\bigcap_n G_n$, and since it does not meet $X\setminus\overline U$, it must meet $\bigcap_n O_n$.  This means that $\bigcap_n O_n$ is dense in $U$.