Set Union Preserves Subsets

Theorem
Let $A, B, S, T$ be sets.

Then:
 * $A \subseteq B, \ S \subseteq T \implies A \cup S \subseteq B \cup T$

Strong Form
Let $\mathbb S$ and $\mathbb T$ be sets of sets.

Suppose that for each $S \in \mathbb S$ there exists a $T \in \mathbb T$ such that $S \subseteq T$.

Then $\bigcup \mathbb S \subseteq \bigcup \mathbb T$.

Proof
Let $x \in \bigcup \mathbb S$.

By the definition of union, there exists an $S \in \mathbb S$ such that $x \in S$.

By the premise, there exists a $T \in \mathbb T$ such that $S \subseteq T$.

By the definition of Subset, $x \in T$.

Thus by the definition of union, $x \in \bigcup \mathbb T$.