Union of Derivatives is Subset of Derivative of Union

Theorem
Let $T = \left({S, \tau}\right)$ be a $T_1$ topological space.

Let $\mathcal F \subseteq 2^S$ be a subset family of $S$.

Then $\displaystyle \bigcup_{A \in \mathcal F} A' \subseteq \left({\bigcup_{A \in \mathcal F} A}\right)'$

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

Proof
Let $\displaystyle x \in \bigcup_{A \in \mathcal F} A'$.

Then by definition of union there exists $A \in \mathcal F$ such that
 * $(1)$: $x \in A'$.

By Set is Subset of Union/Set of Sets
 * $\displaystyle A \subseteq \bigcup_{A \in \mathcal F} A$

Then by Derivative of Subset is Subset of Derivative we have
 * $\displaystyle A' \subseteq \left({\bigcup_{A \in \mathcal F} A}\right)'$.

Hence by $(1)$ the result:$\displaystyle x \in \left({\bigcup_{A \in \mathcal F} A}\right)'$ by definition of subset.