Why empty set? (consequences of Baire's theorem)

Question

I did not understand the proof of Theorem 5.13 of Rudin, [Real and Complex Analysis]. See next. In a complete metric space X which has no isolated points, no countable dense set is a $G_{\delta}$. Proof. Let $\{x_k\}$ be the points of a countable dense set $E$ in $X$. Assume that $E$ is a $G_{\delta}$. Then $E=\cap V_n$, where each $V_n$ is dense and open. Let $W_n=V_n-\cup_{k=1}^n\{x_k\}$. Then each $W_n$ is still dense and open, but $\cap W_n=\emptyset$ (in contradiction to Baire's theorem). I did not understand why $\cap W_n=\emptyset$?

Answer 1

Assume $\cap W_n \neq \emptyset$. Then let $w \in \cap W_n$.

If $\exists k| w = x_k$ then $\forall n\geq k : w \not \in W_n$ so $w \not \in \cap W_n$ (because the $-\cup \{x\}$ term excludes $w$ from such $W_n$), and therefore there is at least one $W_n$ such that $w \not \in W_n$ which would mean $w \not \in \cap W_n$.

Therefore if $w$ exists $w$ cannot be one of the $x_k$, which means that $w \not \in E = \cap V_n$. So $\exists j | w \not \in V_j$.

Now consider for such a $j$ whether $w \in W_j$. On the one hand, $W_j \subset V_j$, on the other hand $w \not \in V_j$ so $w \not \in W_j$. Then $w \not \in \cap W_n$. This contradicts our assumption that there exists a $w \in \cap W_n$; therefore $\cap W_n = \emptyset$.