Let $S=(0,1)\cup(1,2).$
I claim that 2 is the supremum of the set. Let $A$ be the set of all upper bounds of $S$. Since we are dealing with real numbers, and $S$ has an upper bound, I claim that it has least upper bound and that l.u.b is 2. Why is it 2? Let $s \in S$ and let it be the l.u.b. We see that $s < 2$. Because we are in the reals there exists an $r \in \mathbb{R}$ such that $s < r < 2$. Hence $r$ is larger than $s$, so $s$ cannot be another upper bound.
Is this proof on the right track?