Open set as a finite union of closed sets I was studying a text (http://arxiv.org/abs/math/0510331) when in a proof the author stated:
!proof][1 http://www.ime.unicamp.br/~ra151530/mann.png
If I got it right, the author says that $\widetilde{U}$ (a nonempty open connected set in $\mathbb{C}^n$) is a finite union of closed sets, so one of these closed sets must have nonempty interior. 
I tried to prove it but I could not. I think it is a very simple topological detail I am missing.

Suppose for contradiction there are closed sets $F_1, \ldots, F_n$ with empty interior and a nonempty open set $G$ such that $G = F_1 \cup \ldots \cup F_n$. 
Then $G_1 = G \setminus F_1 = G \cap (\mathbb{C}^n \setminus F_1)$ is nonempty (because $F_1$ has empty interior) and open, because it is an intersection of two open sets. 
Again, the set $G_2 = G_1 \setminus F_2$ is also nonempty open. Continue this to deduce $G_n = G_{n-1} \setminus F_n$ is nonempty, which contradicts the assumption $G = F_1 \cup \ldots \cup F_n$. 
