Let $S \subset\mathbb{R}$ and $u$ an upper bound of $S$. If $u \in S$, then $\sup S= u$.
Is my proof correct?
Suppose $u \in S$.
$u$ is an upper bound
Suppose there exists another upper bound $v \in S$. Since $v$ is an upper bound $v>u$, a contradiction. So $v \leq u$.
I am confused since from the definition of supremum, for all upper bound $d$, $d$ must be greater than or equal to $u$, but I have shown otherwise, which is $v \leq u$.
That's not ok. First of all, you didn't prove that $v\leq u$, because if $v$ is an upper bound, then all you really have is $v\geq u$. Second of all, you have no reason to assume that $v\in S$.
What you want to prove is that
There is no demand that $v\in S$. $v$ can be a real number.
So, restart your proof, and start with
"Suppose $v\in\mathbb R$ is an upper bound for $S$"
And you have to finish with $v\geq u$