Here's the proof for why $\sup(a,b) = b$ in AOPS Calculus:
$x < b$ for any $x ∈ (a, b)$, so $b$ is an upper bound.
$x ∈ (a, b)$ for any $x ∈ \Bbb R$ and $a < x < b$, so there is no upper bound less than $b$ $⟹ \sup(a, b) ≥ b \quad(1)$
But $\sup(a,b) ≤ b$ by definition. $\quad(2)$
From $(1), (2)$, we conclude that $\sup(a, b) = b$
I feel there is a logical leap in the line that leads to point (1). It's intuitive, but I wondered if it's possible to prove it more rigorously.
Suppose there is an upper bound $c$ less than $b$, then c must satisfy: $$\begin{cases}c < b \\ c ≥ x \,\, for \,\, all \,\, x ∈ (a, b)\end{cases}$$
This implies that $c ∈ (a,b)$.
Such an upper bound $c$ doesn't exist because we could always find a number $x ∈ (a,b)$ that is greater than $c$. But how do we prove this statement?