I'm working through Kadets' book A Course on Functional Analysis and Measure Theory to brush up on my measure theory and wanted to spend a bit more time with the Baire category theorem. This is Exercise 2.1.2.8 in the book.
Proposition: The intersection of a finite or countable number of dense $G_\delta$ sets in a complete metric space is again a dense $G_\delta$ set.
Proof: Let $X$ be the underlying complete metric space and let $(A_n)_{n=1}^\infty$ be a sequence of dense $G_\delta$ sets. We can write each $A_n$ as:
$$A_n=\bigcap_{k=1}^\infty U_{n,k},$$
where $U_{n,k}$ is open for each pair $(n,k)$. The intersection $A$ of these is:
$$A=\bigcap_{n=1}^\infty A_n=\bigcap_{n=1}^\infty\bigcap_{k=1}^\infty U_{n,k}=\bigcap_{n,k=1}^\infty U_{n,k},$$
a countable intersection of open sets, and thus $A$ is also a $G_\delta$ set.
It remains to show that $A$ is dense in $X$. First note that for each pair $(n,k)$, $U_{n,k}^c$ is nowhere dense in $X$. Indeed, we have:
$$\left(\overline{U_{n,k}^c}\right)^\circ=\left[\left(U_{n,k}^\circ\right)^c\right]^\circ=\left(\overline{U_{n,k}}\right)^c=X^c=\varnothing.$$
On the other hand, suppose that there exists $x_0\in X$ and $r_0>0$ such that $A\cap B(x_0,r_0)=\varnothing$; then for each $r<r_0$, the closed ball $C(x_0,r)$ lies in $A^c$, and we have:
$$C(x_0,r)=C(x_0,r)\cap A^c=C(x_0,r)\cap\bigcup_{n,k=1}^\infty U_{n,k}^c=\bigcup_{n,k=1}^\infty\left[U_{n,k}^c\cap C(x_0,r)\right].$$
Thus $C(x_0,r)$ is a countable union of nowhere dense sets; being a closed subset of a complete metric space, $C(x_0,r)$ is complete, and thus is empty by the Baire category theorem. Since each $x\in B(x_0,r_0)$ lies in $C(x_0,r)$ for some $r<r_0$, it follows that $B(x_0,r_0)$ is in fact empty, in contradiction with the assumption that $r_0>0$. It follows that $A$ is dense in $X$. $\blacksquare$
Best Answer
Your proof seems correct to me. Also you can use an equivalent form of Baire Category Theorem: Countable intersection of dense open sets is again dense. Note that with this form your equation: $$ A = \bigcap_{n,k=1}^\infty U_{n,k} $$ Immediately implies what you want, because the $U_{n,k}$ 's are again countable.