Resolvability is open subspace-closed

Statement
Suppose $$X$$ is a resolvable space and $$U$$ is an open subset of $$X$$. Then, $$U$$ is a resolvable space with the subspace topology.

Facts used

 * 1) uses::Intersection of dense subset with open subset is dense in the open subset

Proof
Given: A topological space $$X$$ with disjoint dense subsets $$C$$ and $$D$$. An open subset $$U$$ of $$X$$.

To prove: $$U$$ has two disjoint dense subsets.

Proof: