Density of Baire’s Theorem for Complete Metric Spaces 
Let $\{G_n\}$ be a sequence of dense open subsets of a complete metric space $X$. Show that $ \bigcap_{n \geq 1}G_n$ is nonempty.

Proof:
Consider some point $p_1 \in G_1$. There exist real numbers $0< r_1 <r<1$ such that 
$$N_1 = N_{r_1} (p_1) \subset N_r (p_1) \subset G_1.$$
Let $q$ be a limit point of $N_1$. For any $\epsilon > 0$ we can find $x \in X$ such that $x \in N_1$ and $x \in N_\epsilon (q)$. Hence 
$$d(p_1, q) \leq d(p_1, x) + d(x, q)<r_1 + \epsilon.$$
Since $\epsilon$ was arbitrary it follows that 
$$d(p_1, q) \leq r_1.$$
Thus, if $q$ is any limit point of $N_1$, $q$ is an interior point of $G_1$ (to see this, simply consider the neighbourhood of radius $0<\epsilon<r-r_1$) and so we can write
$\overline{N_1} \subset G_1.$
Since the $G_n$ are dense, either $p_1$ is a point in $G_2$ or a limit point of $G_2$. In either case, arguing as above, we can find a point $p_2 \in G_2 $ and a neighbourhood of $p_2$ with radius $r_2 < \min\big(r_1, \frac{1}{2}\big)$ such that
$$\overline{N_2} \subset \overline{N_1}$$
and, since the $G_n$ are all open, we also choose $r_2$ small enough that
$$\overline{N_2} \subset G_2.$$
Continuing this process, we can construct the sequence of closed subsets $\Big\{\overline{N_n}\Big\}$. Further, by the way we have chosen the $r_n$ (i.e., $0<r_n<\frac{1}{n}$), it follows $r_n \to 0$, so we have actually constructed a nested sequence of closed bounded sets $\Big\{\overline{N_n}\Big\}$ such that $diam \overline{N_n} \to 0.$ Since $X$ is a complete metric space, the intersection of this sequence is nonempty and since each $$\overline{N_n} \subset G_n$$ the result follows. $\qquad \square$
Assuming this proof is correct, I am wondering if it can be extended to show that the intersection of the $G_n$ is actually dense, or if that requires a completely different line of argument?
 A: That the intersection of the $G_n$ is dense, is a small modification of the above argument: let $O$ be any non-empty open set in $X$, and start with an open ball $N_0 \subseteq O$ and stay inside $N_0$ with all subsequent steps. This hardly takes any effort at all, but does show that the $x \in \cap G_n$ is also in $N_0$ hence in $O$. So the intersection of the $G_n$ intersects every non-empty open set, hence is dense.
The construction of the $N_n$ can be a bit simplified:


*

*Start with $p_0 \in O$ and $N_0 := B(p, r_0) \subseteq O$.

*$N_0$ is open, so $G_1 \cap N_0$ is non-empty (as $G_1$ is dense) and open (as $G_1$ is open). So pick $p_1 \in G_1 \cap N_0$ and let $0< r_1 < 1$ be small enough that $\overline{B(p_1, r_1)} \subseteq G_1 \cap N_0$. Define $N_1 = B(p_1, r_1)$.

*$N_1$ is open and again we have that $p_2 \in N_1 \cap G_0 \cap G_1$ exists by denseness and this set is open so there exists $0<r_2 < \frac12$ such that $\overline{B(p_2, r_2}) \subseteq (N_1 \cap G_0 \cap G_1)$. Define $N_2 = B(p_2, r_2)$.

*continue this process recursively.


No distinguishing limit points etc. Just go straight to the goal.
Then the $\overline{N_n}$ $n \ge 1$ form the required nested family that the Cantor intersection theorem can be applied to. The promised $p \in \bigcap_{n \ge 1} \overline{N_n} \subseteq O \cap \bigcap_{n \ge 1}O_n$ witnesses the denseness of $\bigcap_{n \ge 1} G_n$.
A: The proofs of @Bollo and @Brandsma are great and I try to supplement a situation when the metric space is finite.
Supplement proof:
When the metric space $X$ is a finite set, it is always complete. However, there is no such open set $N_r(p_1) \subset G_1$ (in your proof) because every point is an isolated point. Therefore the only dense open subset of $X$ is itself. So the intersection is itself, which is dense.
