Supremum of Set of Real Numbers is at least Supremum of Subset/Proof 3

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

We know that $\sup T$ and $\sup S$ exist.

So, $\sup T \le \sup S$ exists.

Let us try to evaluate $\sup T \le \sup S$.

Observe that:


 * $t \in S$ for every $t \in T$ since $T \subseteq S$

We find: