I was hoping someone could check my proof.
Thm. Define the open interval $E := (a,b)$ in $\Bbb R$. Prove that $\sup E = b$.
Proof. Since $\emptyset \ne E\subset\Bbb R$ is bounded above (e.g., by $b$), $\exists\sup E$ since $\Bbb R$ possesses the $LUB$ property. Since $b$ is an upper bound, we have $x \leq b\;\forall x\in E$. By the definition of supremum, $\sup E \leq b$. For a contradiction, suppose $\sup E \neq b$. Then consider the element $$\beta = \frac{\sup E + b}{2}.$$ So $a \leqslant \sup E < \beta < b$, so $\beta \in E$. But this implies that $\sup E$ is not an upper bound of $E$, a contradiction. Hence, $\sup E = b$.
Update. I have come across one additional question about thinking about the result more. If $a = b$, $(a,b) = \emptyset$, and $\sup E$ does not exist. Can I begin the proof with "without loss of generality, suppose $a < b$?"
Correct, yes.
To nitpick, I would delete the second sentence; you never use it, and in the first sentence you already asserted (in the parenthetical) that $b$ is an upper bound, so it's implied that you/your reader already know the definition.