Looking for suggestions and verification on this proof.
The exercise reads as follows:
Assume that $A$ and $B$ are nonempty, bounded above, and satisfy $B \subseteq A$. Show $sup B \leq sup A$.
And this is my proof!
Proof:
Say $B = A$. Then $supB=supA$, so certainly $supB \leq supA$.
Say $B$ is a proper subset of $A$. Assume $supB > supA$. Take some element $x = \frac{supB+supA}{2}$. Then since $supB > supA, x \in B$ and $x \not\in A$, since $\frac{supB+supA}{2} > supA$ since $supB > supA$. But then $B \not\subseteq A$ and $B \subseteq A$, which is a contradiction. So the assumption is false.
Something about my proof doesn't feel right, but I can't figure out what.
Supremum definition
$\sup(B)$ is the smallest M such that $\forall x\in B, \ x\leq M \tag 1$
Proof
$\forall x \in B, $ we have $x\in A $ therefore $x\leq \sup(A)$
So $\sup(A)$ is such a M from equation $(1)$ and $\sup(B)$ is the smallest therefore: $$\sup(B)\leq \sup(A) \tag 2$$
About your proof
The only issue with your proof I see is assuming $x = \dfrac{\sup(A) + \sup(B)}{2}$ belongs to B even thought the intuition is there this is not true.
What you can be certain which follows the same logic is that $ \exists x\in B$ s.t $\sup A<x \leq \sup B$ because of the definition of $\sup(B)$ and $\sup(A) <\sup(B)$. So that $x\in B$ and $x\notin A$ and your proof would work. But again the key property that leads to the existence of such a x you want to prove is the minimality of the supremum.