Resolvability is open subspace-closed

This article gives the statement, and possibly proof, of a topological space property (i.e., resolvable space) satisfying a topological space metaproperty (i.e., open subspace-closed property of topological spaces)
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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. 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:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$C \cap U$ and $D \cap U$ are both dense in $U$.	Fact (1)	$C,D$ dense in $X$ $U$ open in $X$		given-fact direct
2	$C \cap U$ and $D \cap U$ are disjoint.		$C,D$ are disjoint		basic set theory!
3	$C \cap U$ and $D \cap U$ are the desired disjoint dense subsets in $U$			Steps (1), (2)	Step-combination direct