Let $(X,\le)$ be a chain, $S\subseteq X$, and suppose $\sigma=\sup S$ exists. Then for every $x\in X$ with $x<\sigma$, there is some $s\in S$ s.t. $x<s\le\sigma$.
By contradiction: Let $x\in X$ s.t. $x<\sigma$ and suppose for every $s\in S,x\not<s$, i.e. $x\ge s$. Then $x$ is an upper bound of $S$ but $x<\sigma$, contradiction.
I want to know if there's a quick alternative direct proof instead of proof by contradiction.
We can phrase just the same proof such that at least it doesn't sound like it's by contradiction:
Because $x<\sigma$, $x$ is not an upper bound for $S$. So there exists $s\in S$ such that $x<s$. However $\sigma$ is an upper bound for $S$, so $s\le\sigma$.
One may nitpick and claim there is a proof-by-contradiction hidden in the step from "not all $s\in S$ are $\le x$" to "there is an $s\in S$ with $s>x$". But in general, worrying unduly about whether this or that proof is by contradiction or not is a bit of a fool's errand unless you're going all the way and explicitly select your definition of concepts such as "chain" and "upper bound" or "supremum" while keeping carefully track of minor difference in the wording that matters intuitionistically, but did not matter to the authors of the textbooks you're otherwise getting those definitions from ...