I have written an answer for the problem 20, chapter 3 of Walter Rudin's Principle of Mathematical Analysis. I think the proof is correct, but since I am new with this kind of proofs, I am skeptical about the subtle points. So it would be really helpful to me if someone can comment on my proof, and maybe give some hints if some error is found. Any help is appreciated. Here is my proof along with the original question and hint:
Exercise 22: Suppose $X$ is a nonempty complete metric space, and $\{G_n\}_{n\ge 1}$ be a sequence of open dense subsets of $X$. Prove Baire's theorem, namely, that $ \bigcap_{n=1}^{\infty}G_n$ is nonempty. (In fact, it is dense in $X$). Hint: Find a shrinking sequence of neighborhoods $E_n$, such that $\overline{E}_n\subset G_n$ and then apply exercise $21$.
Exercise $21$ asks to prove that it $\{E_n\}$ is a decreasing sequence of closed , nonempty and bounded sets in a complete metric space with $\lim_{n\to \infty}\mbox{diam} E_n=0$, then $\bigcap_{1}^\infty E_n$ contains exactly one point.
I have proven exercise 21 and I know that that proof is okay. So, I will show here how I constructed the sequence of $E_n$ for exercise $22$. Also I have tried to prove that $\bigcap_{1}^\infty G_n$ is in fact dense in $X$.
Let $x\in X$. Let $B_{\epsilon}(x)$ be an open ball around $x$.
Claim 1: $\forall n$, $B_{\epsilon}(x)\cap G_n\ne \emptyset$
Proof: If $\exists\ n$ such that $x\in G_n$ then it is obvious. Otherwise, as $G_n$ is dense in $X$, $x$ is a limit point of $G_n$ which implies the result.$\blacksquare$
Now, since each $G_n$ is open and so is $B_{\epsilon}(x)$, $B_{\epsilon}(x)\cap G_n$ is open $\forall n$. Now, let $p_1\in B_{\epsilon}(x)\cap G_n$. Then, $\exists\ \delta>0$ such that $B_{\delta}(p_1)\subset B_{\epsilon}(x)\cap G_1$. Choose $0<\delta_1<\delta$ and set $E_1=B_{\delta_1}(p_1)$.
Claim 2: $\overline{E_1}\subset B_{\delta}(p_1)$
Proof: Clearly, $E_1\subset B_{\delta}(p_1)$. Now, let $q$ be a limit point of $E_1$. Then, $\forall \epsilon'>0,\ \exists p'\in E_1$ such that $d(p',q)<\epsilon'\implies \forall \epsilon'>0,\ d(p_1,q)\le d(p_1,p')+d(p',q)<\delta_1+\epsilon'\implies d(p_1,q)\le \delta_1<\delta$ from which the claim follows.$\blacksquare$
Hence it follows that $\overline{E}_1\subset B_{\epsilon}(x)\cap G_1$.
Let us assume that we have constructed $E_k$ such that $E_1\supset E_2\supset\cdots\supset E_k$ and $\overline{E}_i\subset B_{\epsilon}(x)\cap G_i,\ i=1,2,\cdots,\ k$. Now, $E_k\cap G_{k+1}\ne \emptyset$ and is open by construction. Let $p_{k+1}\in E_k\cap G_{k+1}$, then, $\exists \delta'>0$ such that $B_{\delta'}(p_{k+1})\subset E_k\cap G_{k+1}$. Then, choose $0<\delta_{k+1}<\delta'$ and set $E_{k+1}=B_{\delta_{k+1}}(p_{k+1})\subset E_k\cap G_{k+1}\subset B_{\epsilon}(x)\cap G_{k+1}$, so that $\overline{E}_{k+1}\subset B_{\epsilon}(x)\cap G_{k+1}$. Then, from the Exercise $21$ of the book, it follows that $\bigcap_{1}^{\infty}G_{n}$ is dense in $X$.$\blacksquare$
Best Answer
Your proof is correct.
I would perhaps make the conclusion a bit more explicit. You skip a few steps. They are simple, but an inexperienced reader might be confused. For example, Exercise 21 shows that there exists some point $y\in \bigcap \overline {E_k}$, so there exists some point $y\in \bigcap G_n$ such that $y\in B_\epsilon(x)$. Because $\epsilon$ is arbitrary, every open ball around $x$ contains a point of the intersection $\bigcap G_n$. Since $x$ is arbitrary, this shows $\bigcap G_n$ is dense.