Intersection Dense and Regular Closed Sets is Dense

Question

Let $X$ be a metric space. It seems intuitive to me that if $D$ is dense in $X$ and $K=cl(int(K))$ (ie: $K$ is a regular closed set in $X$) then $$ cl(D\cap K) = K, $$ that is $D$ is dense in $K$....but how to show this/is it actually true?

Answer 1

$int(K) \subset cl(D\cap K)$ because if $x \in int(K)$ and $U$ is any open set containing $X$ then $U\cap int(K)$ is a nonempty open set; so it intersects $D$. Thus every neighbourhood of $x$ intersects $D \cap K$ proving that $int(K) \subset cl(D\cap K)$. Now just take closure on both sides to see that $K \subset cl(D\cap K)$. The reverse inclusion is obvious.