My question reads
If $A$ and $B$ are nonempty, bounded, and satisfy $A\subseteq B$, then $\sup A\le \sup B$.
To me this makes sense because $A$ is a subset of $B$, but I am having issues setting up my proof.
Is it correct to say that because $A$ and $B$ are bounded that they each have a supremum and then use this fact? I was thinking of saying let $a\in A$ be the supremum of $A$ and $b\in B$ be the supremum in $B$. Since $A$ is a subset of $B$, it follows that $a\in\ B$ as well and since $b$ is the supremum of $B$ we have that $a\leq\ b$. Hence, $\sup A\leq \sup B$.
Updated Proof:
Suppose $\operatorname{sup}A>\operatorname{sup}B$. Then, $\operatorname{sup}B$ is not an upper bound for A. Then there exists an $x\in\ A$ such that $x>\operatorname{sup}B$. By subset, $x\in\ B$ where $x>\operatorname{sup}B$, which tells us that $\operatorname{sup}B$ is not an upper bound for $B$. This is a contradiction.
Best Answer
A very direct proof:
Lemma 1: If $w$ is an upper bound of $X$ then $w \ge \sup X$.
Pf: This is almost the definition. It is the contrapositive of definition. If $w < \sup X$ it can't be an upper bound.
Lemma 2: If $A \subset B$ and $w$ is an upper bound of $B$, then $w$ is an upper bound of $A$.
Pf: Immediate consequence of the definitions. For all $a \in A$ then $a \in B$ and $w \ge a$ because it is an upper bound of $B$. So it is an upper bound of $A$.
Now the statement is obvious.
$w = \sup B$ is an upper bound of $B$ $\implies w$ is an upper bound of $A \subset B$ (Lemma 2) $\implies \sup B = w \ge \sup A$ (Lemma 1).