Dense $G_{\delta}$ set implies comeagre set Suppose that $X$ is a metric space. Show that if $D$ is a dense $G_{\delta}$ set, then $D$ is comeagre, that is, countable intersection of dense sets.
My attempt: Let $D=\bigcap_{n \in \mathbb{N}}{D_n}$ where $D_n$ is open. Since $D$ is dense in $X$, then we have $X = \overline{D}=\overline{\bigcap_{n \in \mathbb{N}}{D_n}}=\bigcap_{n \in \mathbb{N}}{\overline{D_n}}$. Hence, $\overline{D_n} =X$ for all $n \in \mathbb{N}$. Therefore, $X \backslash D_n$ is nowhere dense. Since $X \backslash D$ is a countable union of nowhere dense sets, $D$ is comeagre. 
Question: Is $\overline{D_n} = X$ for all $n \in \mathbb{N}$?
 A: The answer to your question is yes, the closure of $D_n$ is $X$.  This is almost immediate: since $D$ is the intersection of the sets $D_n$, we must have $D \subseteq D_n$ for every $n$.  It is a standard fact that for every pair of sets $A,B$, if $A \subseteq B$ then $\overline{A} \subseteq \overline{B}$.  (Note that $\overline{B}$ is a closed set that contains $A$.)  So we must have $X = \overline{D} \subseteq \overline{D_n}$.  The reverse inclusion is obvious, since $\overline{D_n}$ is by definition a subset of $X$.
But there are two other gaps in your proof.

$\overline{\bigcap_{n \in \mathbb{N}}{D_n}}=\bigcap_{n \in \mathbb{N}}{\overline{D_n}}$

This is not true, in general.  (For a counterexample, consider the metric space $\mathbb{Q}$.  Enumerate the rationals as $q_n$ and let $D_n = \mathbb{Q} \setminus \{q_n\}$.  Then the sets $D_n$ are dense (and even open), so $\overline{D_n} = \mathbb{Q}$ and the right side equals $\mathbb{Q}$.  But $\bigcap_{n \in \mathbb{N}} D_n = \emptyset$ so the left side is empty.)

Hence, $\overline{D_n} =X$ for all $n \in \mathbb{N}$. Therefore, $X \backslash D_n$ is nowhere dense.

It is not true for general sets $A$ that if $\overline{A} = X$ (i.e. if $A$ is dense) then $X \setminus A$ is nowhere dense.  (For a counterexample, take $X = \mathbb{R}$ and $A = \mathbb{Q}$.  Double check that you are using the correct definition of "nowhere dense".)   It does happen to be true for open sets, but you need to argue this more carefully.
Once you show this fact (if $A$ is open and dense then $X \setminus A$ is nowhere dense) then the proof is pretty easy.  The sets $D_n$ are open, and above we argued that they are dense, so therefore $X \setminus D_n$ is nowhere dense for each $n$.  By De Morgan's law we have $X \setminus D = \bigcup_{n \in \mathbb{N}} (X \setminus D_n)$ so we see that $X \setminus D$ is meager, meaning that $D$ is comeager.
