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