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...

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.