Proof that a countable intersection of dense open sets in a Baire space is dense

Proof

We pass to complements and use the definition of a Baire space.

1. Let $X$ be a Baire space, and let $U_n$ be open and dense in $X$ for each $n \in \N$. Let $F_n = X \setminus U_n$, a closed set.
2. Fix $n$, and let $V$ be an open set with $V \subseteq F_n$. Then $V \cap U_n = \emptyset$, so $U_n \subseteq X \setminus V$.
3. The set $X \setminus V$ is closed. Since the closure is the smallest closed superset, $\overline{U_n} \subseteq X \setminus V$.
4. Since $U_n$ is dense, $\overline{U_n} = X$. So $X \subseteq X \setminus V$, which gives $V = \emptyset$.
5. The interior $\operatorname{int} F_n$ is an open subset of $F_n$, because the interior is the largest open subset. By steps 2 to 4, $\operatorname{int} F_n = \emptyset$.
6. Since $X$ is a Baire space, $\operatorname{int} \bigcup_n F_n = \emptyset$.
7. Let $D = \bigcap_n U_n$. A point lies outside $D$ if and only if it lies outside some $U_n$. So $X \setminus D = \bigcup_n F_n$.
8. Let $\overline{D}$ be the closure of $D$. By the same statement as in step 3, it is closed and contains $D$. So $X \setminus \overline{D}$ is open.
9. Since $D \subseteq \overline{D}$, we get $X \setminus \overline{D} \subseteq X \setminus D = \bigcup_n F_n$.
10. By steps 8 and 9, and since the interior is the largest open subset, $X \setminus \overline{D} \subseteq \operatorname{int} \bigcup_n F_n = \emptyset$.
11. So $\overline{D} = X$, and $\bigcap_n U_n$ is dense in $X$.