The following is problem 3.22 from Rudin's Princples of Mathematical Analysis:
Suppose $X$ is a nonempty complete metric space, and $\{G_n\}$ is a sequence of dense open subsets of $X$. Prove Baire's theorem, namely, that $\bigcap_{n=1}^\infty G_n$ is not empty. Hint: find a shrinking sequence of neighbourhoods $E_n$ such that $\overline{E}_n\subset G_n$.
Here's what I've tried so far:
Let $\{r_n\}$ be a Cauchy sequence of positive real numbers converging to $0$. Fix $x\in X$ and define $E_i=\{g\in G_i:d(g,x)
, which is nonempty since $G_i$ is dense. I would like to show that for all $i$, $\overline{E}_i\subset G_i$ (I had convinced myself that this would be true but I am now having doubts). Let $e\in \overline{E}_i$. Then either $e\in E_i$ or $e$ is a limit point of $E_i$. If $e\in E_i$ then $e\in G_i$. Otherwise, every neighbourhood of $e$ contains a point in $E_i$.
I thought that I should be able to then choose some point $e'\in E_i$ in a neighbourhood of $e$ and, since $G_i$ is open, it'll have a neighbourhood $N\subset G_i$ which contains $e$, but this is proving to be difficult and I'm worried that it's not true. If I can show that this is true then the rest will follow from results I've already proven.
Does my approach make any sense?
Incidentally, as a secondary question, what type of a thing would $G_n$ be? A sequence of dense open subsets seems weird to me—at first I was thinking of some sequence of infinite subsets of rational numbers in the real numbers but I realized that those aren't open. Is there anything which would be familiar to my little undergrad brain which would be analogous to this problem?