Intersection of two open dense sets is dense

Solution 1:

You can't proceed further from that point since the inclusion you get at the end doesn't give you anything. I suggest a direct approach. To show that $G\cap H$ is dense in $X$, it suffices to show that $G\cap H$ intersects every nonempty open subset of $X$. To this end, let $V$ be an open subset of $X$. Since $G$ is open and dense in $X$, $G \cap V$ is a nonempty open set. Then since $H$ is dense in $X$, $H\cap G \cap V$ is nonempty, as desired.

Solution 2:

Working with interiors and closures is messy; it’s easier to use the fact that a set $D$ is dense in $X$ if $D\cap U\ne\varnothing$ whenever $U$ is a non-empty open set in $X$. Thus, you want to show that if $U$ is a non-empty open set in $X$, then $U\cap(G\cap H)\ne\varnothing$. HINT: $U\cap G$ is a non-empty open set (why?), and $H$ is dense in $X$, so ... ?