## Dense Set Problem

Suppose $(X, \Omega)$ is a topological space, and suppose that $U \subseteq X$ is an open subspace of $X$. If $D$ is a dense subset of $X$, prove that $D \cap U$ is dense in $U$.

Here's my attempt at a proof (with help from other people)... I'm not certain if the argument is sound or not, or even if there is a better way to prove it. We decided to try subset inclusion...

Proof:
$U \subseteq \overline{U \cap D}$

Suppose that the opposite is true, and let $V = U - \overline{U \cap D}$. The set $V$ is open, hence if it is nonempty it must meet the dense set $D$. This is a contradiction, since $V \subseteq U$ and so $V \cap D \subseteq U \cap D$, but $V$ is disjoint from $U \cap D$. Thus, $U \subseteq \overline{U \cap D}$.

(I'm not completely sold on this argument... But I can't think of one, and they assure me it's right... I just don't know why).

$\overline{U \cap D} \subseteq U$

We know that if a set is closed in $U$, then it must equal $C \cap U$ for some $C \in \kappa (\Omega)$. Threfore, $\overline{U \cap D} = C \cap U \subseteq U$.

Any help is greatly appreciated.