I am not sure if my proof is proper, so please comment on it and try to fix it if you can.
Let $S$ be a non-empty set. Say that a and b are both supremum of $S$ with $a<b$.
Assuming the definition of a supremum that it must be the least upper bound of the set.
Proof by contradiction that there is a unique supremum:
Because $b=\sup S$ then $b \ge x$ for all $x \in S$. Similarly because $a=\sup S$ then $a \ge x$ for all $x \in S$
Results:
$b>a \ge x$, this implies $b> x$, but that means $b$ is not a supremum anymore, and $a$ is the new supremum as $a≥ x$. Therefore there must be only one supremum.
We can easily prove the second subpoint of the supremum definition.( For all $r>0$ there exists $m \in S$ : $a-r<m$)
Best Answer
No, that argument doesn't work. It is quite possible for the least upper bound of a set to not be equal to any element - for example, the negative real numbers.
The way to prove this is that word "least" in the name. By definition, $a$ is a least upper bound for a set $S$ if it is an upper bound for $S$ (for all $x\in S$, $x\le a$) such that, for any $b$ that is also an upper bound for $S$, $a\le b$. This definition applies for any order, or even partial order; the order on the real numbers is just one case. Whether least upper bounds exist in general depends on the system we're in.
Oh, and that proof I alluded to? If $a$ and $b$ are both least upper bounds for $S$, then by the definition, $a\le b$ and $b\le a$. Therefore, $a=b$. It's that simple. Don't overcomplicate it.