Supremum of Set of Real Numbers is at least Supremum of Subset/Proof 4

Proof
By definition $\sup S$ is an upper bound for $S$.

Thus:
 * $\forall x \in S: x \le \sup S$

As $T \subseteq S$ we have by definition of subset that:
 * $\forall x \in T: x \in S$

Hence:
 * $\forall x \in T: x \le \sup S$

So by definition $\sup S$ is an upper bound for $T$.

So $\sup S$ is at least as big as the smallest upper bound for $T$

Thus by definition of supremum:
 * $\sup T \le \sup S$