The complement of the double countable union nowhere dense closed sets. I am working on an exercise of functional analysis, which has been partially discussed here: Condensation of Singularities, Principle of Uniform Boundedness and here: The Principle of Condensation of Singularities.
I have finished most of it, but to conclude, I got stuck in topology. The following describes my question.
Say, I have proved that $Int(F_{k,n})=\varnothing$ for all $k,n\in\mathbb{N}$ and $F_{k,n}$ is closed for all $k,n\in\mathbb{N}$. Consider the set defined by $$H_{k}:=\bigcup_{n=1}^{\infty}F_{k,n}\ \ \text{and}\ \ O:=\bigcup_{k=1}^{\infty}H_{k}.$$
The space we are working with is a Banach space $V$, and the set I am considering is $E$ that satisfies $E=V\setminus O$.
I want to use  Baire category theorem to say that $E$ is dense in $V$.
So I start to argue that $O=\bigcup_{k,n=1}^{\infty}F_{k,n}$ is a countable union of closed dense sets, so that $E$ is a countable intersection of open dense sets. Therefore, the Baire category theorem implies that $E$ is dense.
However, I got confused about why $O$ is a countable union of closed dense sets. If we separate the double countable union, we will have $$O=\bigcup_{k=1}^{\infty}\Bigg(\bigcup_{n=1}^{\infty}F_{k,n}\Bigg),$$ but $\bigcup_{n=1}^{\infty}F_{k,n}$ is not necessarily closed right? And in fact $H_{k}:=\bigcup_{n=1}^{\infty}F_{k,n}$ is not necessarily dense anymore. So is this still safe to say $O$ is a countable union of closed dense sets?
Perhaps I am overthinking this and I believe there must be a way to prove the density of $E$ in $V$. Please let me know if you think I should provide more functional setup of this question. But for now I want to keep it as concrete as possible.
Thank you so much!

The collection $\{F_{k,n}:k, n \geq 1\}$ is a countable collection of closed nowhere dense sets and $O$ is its union. (You can always arrange this family in a single sequence $(C_n)$).
