I know there's proof that the irrational number set is a Baire space, but when I tried to prove this, I used a different way. However, I feel my proof is not right.
Let $I= \mathbb{R} \backslash \mathbb{Q}$， I want to prove it is a Baire space. I tried to show that if $\left\{ B_{n} \right\}$ is a family of closed set in $I$ with empty interior, then $$\begin{align*} \bigcup\limits_{n} B_{n} \end{align*}$$ still has empty interior in $I$
Consider any closed set in $I$, it should be in the form $I \cap F$ where $F$ is closed in $\mathbb{R}$. Then it has the empty interior. That is, $I \cap F$ does not contain any open set in $\mathbb{R}$. This is since $\mathbb{Q}$ is dense in $\mathbb{R}$, any open set in $\mathbb{R}$ intersects with $\mathbb{Q}$ nontrivially, so if $I \cap F$ does contain an open set in $\mathbb{R}$, it must contain some rational numbers. This is a contradiction.
Consider let $B_{n}= I \cap F_{n}$ where $F_{n}$ is a closed set in $\mathbb{R}$ , then I think $$\begin{align*} \bigcup\limits B_{n} \subset I \end{align*}$$it shouldn't contain any open set in $\mathbb{R}$ . This is still since $\mathbb{Q}$ is dense in $\mathbb{R}$, so if the union contains any open set of $\mathbb{R}$, then it must contain some rational numbers which contradict with $\bigcup\limits B_{n} \subset I$.
I doubt the correctness of my proof. Is this the right proof? If this is wrong, where it goes wrong? Thanks.