Using the Nested Sphere theorem too Let $(X,d)$ is a complete metric space. Use the Nested Spehere theorem to show that any countable intersection of open dense sets in $X$ is not empty.
Let $\mathcal{U}=\{U \subset X:\ U \ \text{is open dense sets in }\ X \}.$ Let $U_1,U_2,U_3,...$ be any elements in $\mathcal{U}$. Since for each $n \in \mathbb{Z^+},$ $\bar{U_n}=X$, then for any $x \in U_k \subset X$ and any $\epsilon>0$ we have that $$\emptyset \neq B_\epsilon(x) \cap  U_m$$
for any $m$ positive integer. So for $\epsilon_0>0$ such that $x \in B_{\epsilon_0}(x) \subseteq U_k$, we have  So $$\emptyset \neq B_{\epsilon_0}\cap U_m \subseteq U_k\cap U_m  $$ for any $k,m$ positive integer. Does this guarantee that $\displaystyle \bigcap_{n=1}^\infty U_n \neq \emptyset$? Also I am getting stack with the part regards using the Nested Sphere theorem.   
Thanks for any help!
 A: The idea is of course to find a sequence of balls $B_{\epsilon_n}(x_n)\subseteq B_{\epsilon_{n-1}}(x_{n-1})\subseteq...\subseteq B_{\epsilon_1}(x_1)$ with $\epsilon_n\to0$ so that $B_{\epsilon_n}(x_n)\subseteq \bigcap_{k=1}^nU_k$. By the nested ball theorem the sequence $x_1,...,x_n$ converges to some $x$. This $x$ necessarily lies in every $B_{\epsilon_n}(x_n)$, hence in every intersection $\bigcap_{k=1}^n U_k$, hence in $\bigcap_{k=1}^\infty U_k$.
How can you find such a sequence? Start with an arbitrary $x_1\in U_1$. Since $U_1$ is open is a radius $\epsilon_1$ so that $B_{\epsilon_1}(x_1)\subseteq U_1$. Further since $U_1, U_2$ are open dense so is $U_1\cap U_2$ and $B_{\epsilon_1}(x_1)\cap (U_1\cap U_2)$ is non-empty. In particular there is some point $x_2$ in that intersection and some radius $\epsilon_2$ so that that $B_{\epsilon_2}(x_2)\subseteq B_1(x_1)\cap (U_1\cap U_2)$. By making $\epsilon_2$ smaller we can assume it to be $≤\epsilon_1/2$.
That last step was actually an induction argument. Given an $x_n$ and an $\epsilon_n$ we know that $B_{\epsilon_n}(x_n)\cap(U_1\cap...\cap U_{n+1})$ must be non-empty and open, so we can find some $x_{n+1}$ and $\epsilon_{n+1}$ so that $B_{\epsilon_{n+1}}(x_{n+1}) \subseteq B_{\epsilon_n}(x_n)\cap ( U_1\cap...\cap U_{n+1})$ and by making $\epsilon_{n+1}$ smaller we may assume $\epsilon_{n+1}≤\epsilon_n/2$.
These sequences $x_n, \epsilon_n$ then satisfy the conditions that we were looking for.
