Set Union Preserves Subsets

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

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

Proof
Let $A \subseteq B$ and $S \subseteq T$.

Then:

Now we invoke the Constructive Dilemma of propositional logic:
 * $p \implies q, \ r \implies s \vdash p \lor r \implies q \lor s$

applying it as:
 * $\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \lor x \in S \implies x \in B \lor x \in T}\right)$

The result follows directly from the definition of set union:
 * $\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \cup S \implies x \in B \cup T}\right)$

and from the definition of subset:
 * $A \subseteq B, \ S \subseteq T \implies A \cup S \subseteq B \cup T$

Indexed Family of Sets
Let $S_i$ and $T_i$ be indexed families of sets.

Suppose that for each $i \in I$, $S_i \subseteq T_i$.

Then $\bigcup_{i\in I} S_i \subseteq \bigcup_{i \in I} T_i$.

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, $x \in \bigcup \mathbb T$.