prove that a set a can have at most one supremum.
My answer is:
To prove the claim consider the case that there are two (or more) supremum $a$ and $a′$ of a set $A$. Then for every element $x$ of $A$:
$x≤a$ (from the definition of supremum) & $x≤a′$ (from the definition of supremum).
So $a′≤a$(a′ is the supremum of A ) and $a≤a′$(a is the supremum of A).
Now by antisymmetry $a′=a$.
Am I correct?
No, your proof assumes that the supremum is an element of the set. Note that if you have two supremum then both of them are upper bounds and both of them are the least upper bound. Since the first one is the least upper bound and the second one is an upper bound, the the first one is less than or equal the second one. Similarly the second one is less than or equal to the first one,
Thus they are equal and as the result the supremum is unique.