= Solution
A real number $s$ is the <supremum> of a nonempty set $A\subset\mathbb R$ if every $a\in A$ satisfies $a\le s$ and no smaller real number is an <upper bound>. Equivalently, for every $\eta>0$ there is $a\in A$ with $s-\eta<a\le s$. A <greatest element>, or maximum, additionally belongs to $A$.
Let $s=\sup A$ and suppose $s\notin A$. Fix $\varepsilon>0$. The approximation property gives at least one element of $A$ in $(s-\varepsilon,s)$. If there were only finitely many, they would have a largest value $m<s$. Every other element of $A$ is at most $s-\varepsilon$, so $\max\{m,s-\varepsilon\}<s$ would be an <upper bound> for the whole set. This contradicts the definition of the <supremum>. Therefore \b[there are infinitely many such elements], the <accumulation below an unattained supremum>.
For a counterexample when the <supremum> is attained, take \b[$A=\{0\}$]. Its <supremum> and <greatest element> are both zero, while every interval $(-\varepsilon,0)$ contains no elements of $A$.
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