Show that if every point $x \in X$ has a neighborhood that is Baire space, then $X$ is a Baire space. (Munkres "Topology", 48.3)
Here is what I tried :
Let $\{U_n\}_{n \geq 1}$ be a collection of open dense subset of $X$. I would like to show that $\bigcap\limits_{n \geq 1}U_n$ is dense in $X$.
First, let $x$ be an element of $X$ and $V_x$ a neighborhood that is Baire space. Then $\{U_n \cap V_x\}_{n \geq 1}$ is a collection of open dense subsets of $V_x$. Since $V_x$ is a Baire spce, we have that $\bigcap\limits_{n \in \geq 1}(U_n\cap V_x)$ is dense in $V_x$.
In particular, $x$ is a limit point of $$\bigcap\limits_{n \in \geq 1}(U_n\cap V_x) = \left(\bigcap\limits_{n \geq 1}U_n\right)\cap V_x .$$
So $x$ is a limit point of $\bigcap\limits_{n \geq 1}U_n$. But $x$ is any element of $X$ so $\bigcap\limits_{n \geq 1}U_n$ is dense in $X$.
Is this argument valid ?