In Apostol's Mathematical Analysis the following theorem's proof was left as an exercise:
Theorem 1.16 (Comparison Property). $$ \text{Given nonempty subsets S and T of } \mathbb{R} $$ $$ \text{ such that s ≤ t for every s in S and t in T, } $$ $$ \text{ if T has a supremum then S has a supremum and } $$ $$ \sup S \leq \sup T. $$
I would like to know if my following proof is valid,
Proof. For every s in S and t in T we have, $$ s \leq t$$ so S is bounded above by T. Also S is a nonempty subset of $\mathbb{R}$ so by the Completeness Axiom $$\exists \sup S$$ and $\sup T$ is already given. Now, $$s \leq t \leq \sup T \quad \forall s \in S \ \forall t \in T$$ $$ \Rightarrow \sup T \in T$$ and $$s \leq \sup S \quad \forall s \in S$$ $$ \Rightarrow \sup S \in T.$$ Assume for contradiction that $$ \sup S > \sup T. $$ Then $\sup S$ is an upper bound of T but not the least one, so $$t \leq \sup T < \sup S \quad \forall t \in T$$ $$\Rightarrow t < \sup S \quad \forall t \in T$$ $$\Rightarrow \sup S \notin T$$ which is a contradiction as by definition of the supremum $$s \leq \sup S \quad \forall s \in S$$ $$ \Rightarrow \sup S \in T.$$ Thus on negating our assumption we get $\sup S \leq \sup T. \blacksquare$