Supremum of Set of Real Numbers is at least Supremum of Subset/Proof 2

Theorem

Let $S$ be a set of real numbers.

Let $S$ have a supremum.

Let $T$ be a non-empty subset of $S$.

Then $\sup T$ exists and:

$\sup T \le \sup S$

Proof

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

It follows from Supremum of Subset that $\sup T \le \sup S$.

$\blacksquare$