Union of Subsets is Subset/Proof 2

Theorem
Let $S_1$, $S_2$, and $T$ be sets such that:

Suppose that $S_1$ and $S_2$ are both subsets of $T$.

Then $S_1 \cup S_2 \subseteq T$.

That is:
 * $\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right) \implies \left({S_1 \cup S_2}\right) \subseteq T$

Proof
Let $x \in S_1 \cup S_2$.

By the definition of union, either $x \in S_1$ or $x \in S_2$.

By hypothesis, $S_1 \subseteq T$ and $S_2 \subseteq T$.

By definition of subset:
 * $x \in S_1 \implies x \in T$
 * $x \in S_2 \implies x \in T$

By Proof by Cases it follows that $x \in T$.

Hence the result by definition of subset.