An upper bound $u$ of a nonempty set $S$ in $\mathbb{R}$ is the supremum of $S$ if and only if for every $\epsilon>0$ there exists an $s_{\epsilon}\in S$ such that $u-\epsilon<s_{\epsilon}$
Question
The consequence of this lemma could derive from $u-\epsilon<s_{\epsilon}\implies u\leqslant s_{\epsilon}$ for some $s_{\epsilon}$ if $u$ is the least upper bound. But this only hold for such least upper bound is an element of $S$.
This seems may not hold if $\sup A\notin S$.
This was discovered from I was trying to show $\max A=\sup A$
re-edit: In the capture has shown for every $a\in A$, $a\leqslant \max A\leqslant \sup A$ by definition. We also know the consequence of the lemma that for every $\epsilon>0$, there exists $s_{\epsilon}\in A$ such that $\sup A-\epsilon<s_{\epsilon}\leqslant \max A$ for a fixed $\max A$. Hence, $\sup A\leqslant \max A$
Your error lies in forgetting that the $s_{\epsilon}$ depends on $\epsilon$, as the notation itself suggests. As $\epsilon$ changes, $s_{\epsilon}$ may change as well. Think about the set $\{1-\frac{1}{n}\}$ and $u=1$.
It is true that for a fixed $a$, if for every $\epsilon\gt0$ we have $u-\epsilon\lt a$, then we can conclude that $u\leq a$. But here you do not have a fixed $s_{\epsilon}$, so the jump to "$u\leq s_{\epsilon}$" is invalid.
Remember "for every a there is a b" is not the same as "there is a b that works for every a." The implication you want holds in the latter case, but you only have the former here.
It would be less confusing for you if you phrased the property more suggestively. Instead of the dangling clause "for some $s_{\epsilon}$" at the end, with an unclear scope, phrase it as in the statement you quote: "for every $\epsilon$ there exists $s_{\epsilon}$ such that..."