Supremum of Subset of Bounded Above Set of Real Numbers

Theorem
Let $A$ and $B$ be sets of real numbers such that $A \subseteq B$.

Let $B$ be bounded above.

Then:
 * $\sup A \le \sup B$

where $\sup$ denotes the supremum.

Proof
Let $B$ be bounded above.

By the Continuum Property, $B$ admits a supremum.

By Subset of Bounded Above Set is Bounded Above, $A$ is also bounded above.

Hence also by the Continuum Property, $A$ also admits a supremum.

$\sup A > \sup B$.

Then:
 * $\exists y \in A: y > \sup B$

Thus by definition of supremum, $y \notin B$.

That is:
 * $A \nsubseteq B$

which contradicts our initial assumption that $A \subseteq B$.

Hence the result by Proof by Contradiction.