Union of Set of Dense-in-itself Sets is Dense-in-itself

Theorem
Let $T$ be a $T_1$ topological space.

Let $\mathcal F \subseteq \mathcal P \left({T}\right)$ such that
 * every element of $\mathcal F$ is dense-in-itself.

Then the union $\bigcup \mathcal F$ is also dense-in-itself.

Proof
By Dense-in-itself iff Subset of Derivative:
 * $\forall A \in \mathcal F: A \subseteq A'$

Then by Set Union Preserves Subsets:
 * $\displaystyle \bigcup \mathcal F \subseteq \bigcup_{A \in \mathcal F} A'$

where $A'$ denotes the derivative of $A$.

By Union of Derivatives is Subset of Derivative of Union:
 * $\displaystyle \bigcup_{A \in \mathcal F} A' \subseteq \left({\bigcup \mathcal F}\right)'$

Then by Subset Relation is Transitive:
 * $\displaystyle \bigcup \mathcal F \subseteq \left({\bigcup \mathcal F}\right)'$

The result follows by Dense-in-itself iff Subset of Derivative.