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

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

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

(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).

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

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

Any help is greatly appreciated.