Proving that a supremum is unique by contradiction 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 $m \in S$ : $a-r<m$)
 A: 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.
A: In partial order, two elements may not be comparable. Hence we can say that a set may have multiple supremums that are not comparable, so the supremum is not unique with the partial order (in generality). But if both elements of our space can be compared (that is, if the order is general), we will prove below that the supremum is unique. Let $a$ and $b$ both the supremum of $A$
From the definition of supremum, $a$ and $b$ are upper bounds of $A$.If $a$ is a supremum of $A$, then $a \le b$. Because of $b$ is an upper bound of $A$. If $b$ is a supremum of $A$, then $b \le a$. Because of $a$ is an upper bound of $A$. Hence by the antisymmetry of the ordering we obtain$a=b$.
