The intersection of a regular open set $U$ with a dense set $D$ is a regular open set in $D$ If $U$ is a regular open set in $X$, ($U=Int(\overline{U})$) and $D$ is a dense subset of $X$, then $V=U\cap D$ is a regular open set in $D$.
 A: The main challenge here is keeping the
topologies, closures and interiors straight.
Let $\tau_D, \tau$ be the classes of open sets in $D, X$ respectively.
The open sets of $D$ are sets of the form $D \cap O$, where $O \in \tau$.
For want of better notation,let $\overline{\tau_D}$, $\overline{\tau}$ be the classes of closed sets in $D, X$ respectively.
The closed sets in $D$ are sets of the form $D \cap C$, where $C \in \overline{\tau}$.
Let $\overline{\cdot}^D, \overline{\cdot}$ represent closure with respect to
$\overline{\tau_D}, \overline{\tau}$ respectively,
let $\cdot^{\circ_D}, \cdot^{\circ}$ represent the interior
with respect to
$\tau_D, \tau$ respectively.
Since $D$ is dense in $X$, if $D \subset C$, with $C\in \overline{\tau}$, then $C=X$.
A relevant fact about dense sets:
If
$O \in \tau$, and $O \cap D \subset C\in \overline{\tau}$, then $O \subset C$. To see this, note
that $D \subset C \cup (X \setminus O)$, and since the latter set is closed, we have $X = C \cup (X \setminus O)$, from which it follows that $O \subset C$.
We have $U = (\overline{U})^\circ$.
First I claim that $\overline{V}^D = \overline{U} \cap D$. The containment
$\overline{V}^D \subset \overline{U} \cap D$ is immediate. Since $\overline{V}^D \in \overline{\tau_D}$, we have
$\overline{V}^D = C \cap D$ for some $C \in \overline{\tau}$. Since
$V= U \cap D \subset \overline{V}^D$, we have $U \cap D \subset C$, and since $U \in \tau$, we have $U \subset C$ and since $C \in \overline{\tau}$, we have $\overline{U} \subset C$. Consequently, $\overline{U} \cap D \subset C \cap D$.
Now I claim that $(\overline{V}^D)^{\circ_D} = U \cap D$. Since
$U \cap D \in \tau_D$, it follows that $(\overline{V}^D)^{\circ_D} \supset U \cap D$. Now suppose $O \subset \overline{V}^D = \overline{U} \cap D$, with $O \in \tau_D$. We can write $O= O' \cap D$, with $O' \in \tau$, and
this gives $O' \cap D \subset \overline{U} \cap D \subset \overline{U}$. Since $\overline{U} \in \overline{\tau_D}$, we have $O' \subset \overline{U}$
and so $O' \subset (\overline{U})^\circ = U$. Consequently,
$O \subset U \cap D$.
A: Ok I'll try to answer myself.
Denote by $V=U\cap D$, what we want to prove is that $Int_D(\overline{V}^D)=V$. We also note that $\overline{V}^D=\overline{V}\cap D$, (the closure of $V$ in $D$.) and $\overline{V}=\overline{V}^X$.
First we are going to prove the following: $Int_D(\overline{U}\cap D)= Int(\overline{U})\cap D$.
Let $W=Int_D(\overline{U}\cap D)$. If we remember that $W$ is the biggest open set in $D$ contained in $\overline{U}\cap D$, then $U\cap D\subseteq W \subseteq \overline{U}\cap D$. Now suppose that $W$ is necessarily bigger that $U\cap D$, that is; $U\cap D\subsetneq W \subseteq \overline{U}\cap D$, and let $W'$ be an open set in $X$ such that $W=W'\cap D$.
Then by the density of $D$, if we recall this elemental result: For every open set $H$, $\overline{H\cap D}=\overline{H}$. Taking closures in the previous sequence of contentions we get:
$$\overline{U\cap D}\subseteq \overline{W'\cap D}\subseteq \overline{U}\Longrightarrow \overline{U}\subseteq \overline{W'}\subseteq \overline{U}$$
By the previous, $\overline{U}=\overline{W'}$, and by the regularity of $U$ we get, $U=Int(\overline{U})=Int(\overline{W'})$. Now since $W'$ is open in $X$, $W'\subseteq Int(\overline{W'})=U$. So contradicting the hypothesis "$U\cap D\subsetneq W \subseteq \overline{U}\cap D$". Then $Int_D(\overline{U}\cap D)=W=U\cap D= Int(\overline{U})\cap D$.
Finally,
\begin{align*}
Int_D(\overline{V}^D)=Int_D(\overline{V}\cap D)&=Int_D(\overline{U\cap D}\cap D)\\
&=Int_D(\overline{U}\cap D)\\
&=Int(\overline{U})\cap D\\
&=U\cap D=V.
\end{align*}
That's my proof, please if someine find mistake let me notice. :3.
