In the statement of the Baire Category Theorem, one needs to include openness of a set so that the theorem holds.
Question: what is the example such that countable intersection of dense sets is not dense?
EDIT: perhaps I should rephrase my question:
Give an example such that countable intersection of dense sets is not dense and nonempty.
On
Give an example such that countable intersection of dense sets is not dense and nonempty.
It's easy to modify the given counterexample of $\Bbb Q$ and $\Bbb R \backslash \Bbb Q$ to match your needs. For example, you can take,
$$ \Bbb Q \cup \{\pi\}, \ \Bbb Q \cup \{\pi, \pi^2\}, \ \Bbb Q \cup \{\pi, \pi^2, \pi^3 \}, \ \dots, $$
these sets are all dense in $\Bbb R,$ but their intersection in $\{\pi\}.$
In fact, you can prove a stronger result, namely that there exists a countable collection of dense subsets of $\Bbb R$ whose intersection is pairwise not dense and non-empty. Indeed, if $p_1, p_2, \dots$ is an enumeration of the primes, then the collection, $$\ (\Bbb Q + \sqrt{p_1}) \cup \{\pi\}, \ (\Bbb Q + \sqrt{p_2}) \cup \{\pi\}, \ (\Bbb Q + \sqrt{p_3}) \cup \{\pi\}, \ \dots, $$ where $\Bbb Q + \sqrt{p_i} = \{ \sqrt{p_i} + q \mid q \in \Bbb Q \}.$ It can be shown that these sets are dense in $\Bbb R,$ but the intersection of any subcollection is $\{\pi\}$ (this isn't exactly easy to prove, but see this post for one).
Even a finite intersection need not be dense. In the reals, $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are both dense, and have empty intersection.
If we have finitely many dense and open subsets $D_1,\ldots,D_N$, in any space $D_1 \cap \ldots \cap D_N$ is (open) and dense. This is a motivation for Baire spaces: here we can go to countable intersections, not just finite ones. The intersection need not be open anymore, but is still dense in Baire spaces.
To get a non-empty intersection of two, add $(0,1)$ to both the rationals and the irrationals. The intersection is $(0,1)$, not dense.