Prove if sup $S \in S \Rightarrow$ sup $S =$ max $S$ This is might be too easy, but just to make sure I am on the right track.

Prove that if sup $S \in S \Rightarrow$ sup $S =$  max $S$

By definition, the maximum $max$ of a set $S$ is the number that is greater or equal to all the elements of $S$. The supremum $sup$, if in the set, must be greater or equal to all numbers in the set, therefore $sup S = max S$.
Is this correct? Am I missing something?
 A: I think you need to prove it rigorously.
$x\leq $ sup $S$ for all $x\in S$. Also $x\leq$ max $S$ for all $x\in S$. 
Since sup $S\in S, $ sup $S\leq$ max $S$. Since always max $S\in S,$ max $S\leq $ sup $S$.
So, max $S=$sup $S$ 
A: It's certainly correct, technically. Stylistically, it feels to me like it's been in the reverse of the logical order. That is to say, the logic seems to flow better as:

The supremum is defined to be greater than or equal to every element in $S$. Given that $s=\sup S$ is a member of $S$, it is thus an element of $S$ greater than or equal to any given element of $S$. This is the definition of the maximum, hence $s$ is the maximum of $S$ as well.

We're trying to conclude something about the maximum given something about the supremum, so it makes sense to start by talking about the supremum.
A: We can prove the reverse as well, that is, if sup S = max S ⇒ sup S ∈ S.
Assume the opposite, that is, sup S ∉ S.
Hence, x < sup S for all x ∈ S.
Also, x ≤ max S for all x ∈ S.
As max S ∈ S always,
Hence, max S < sup S.
This contradicts our assumption that sup S = max S.
Hence, our assumption was wrong.
Hence, sup  S ∈ S.
Hence,  sup S = max S ⇒ sup S ∈ S.
So finally,
sup S ∈ S ⇔ sup S = max S.
