Let $S$ and $T$ be nonempty bounded subsets of $\mathbb R$ with $S \subseteq T$. Prove that $$\inf T \le \inf S \le \sup S \le \sup T.$$
Proof:
Since $S$ is bounded, then $S$ is bounded above and bounded below.
So $m=\sup S$ and $n=\inf S$ where $m \ge n$.
By the same argument above, $a = \sup T$ and $b= \inf T$ where $a \ge b$.
Since $S \subseteq T$, $\sup T$ is an upper bound for $S$ and $\inf T$ is a lower bound for $S$, so $a \ge m \ge n \ge b$.
Is my proof correct? And if it is, can it be written better?
I think you pretty much have it. I think if I had written the proof, I would have talked about $T$ first and then transitioned into the definition of a subset since $S$ is contained in $T$, but that is just personal preference.