Let $X$ be a perfect Polish Space. Prove that every comeager contains an uncountable dense $G_\delta$ set.
It's known that every perfect Polish Space has cardinality $2^{\aleph_0}$. It's easy to see that every comeager contains a $G_\delta$ set, in particular it contains the intersection of a countable family of dense open sets. Since the Baire category theorem, this $G_\delta$ set is still dense. I hope that this particular $G_\delta$ set has cardinality $2^{\aleph_0}$.
As you mentioned, there are open dense sets $U_0, U_1, ...$ such that $\bigcap_{n \in \omega} U_n$ is contained in the comeager set $A$.
Using perfectness, we can construct inductively open sets $V_s$, where $s \in \{0,1\}^{<\omega}$ with the following properties:
$V_s \subset U_{\text{lh }s}$ for any $s \in \{0,1\}^{<\omega}$
$\overline{V_{s\cdot i}} \subset V_s$ for any $s \in \{0,1\}^{<\omega}$ and $i \in \{0,1\}$
$\text{diam }V_s < \frac{1}{\text{lh } s}$ for any $s \in \{0,1\}^{<\omega}$
$V_{s \cdot 0} \cap V_{s \cdot 1} = \emptyset$ for any $s \in \{0,1\}^{<\omega}$.
We then define $$C = \bigcup_{t \in \omega^\omega} \bigcap_{n \in \omega} V_{t|n}.$$
Then $C$ is a Cantor set contained in the comeager set $A$ and since $C$ is uncountable, $A$ is uncountable as well.