From Calculus by Tom Apostol:
DEFINITION OF LEAST UPPER BOUND. A number $B$ is called a least Upper bound of a nonempty set $S$ if $B$ has the following two properties: (a) $B$ is an Upper bound for $S$. (b) No number less than $B$ is an Upper bound for $S$
THEOREM 1.32. Let $h$ be a given positive number and let $S$ be a set of real numbers. If $S$ has a supremum, then for some $x$ in $S$ we have $x\gt\ \sup S\space -h$
Let be the set $A=\{1\}$ $A$ is bounded above and is nonempty, therefore it has least upper bound. The least upper bound is $1$.
Now, let $h = 0,1$ $SupA - h = 0,9$ but $0,9 \notin A$ contradicting the theorem.
In the book, there is no asumption that the set should contain infinite elements or to be dense.
Can you clarify this confusion? Thanks.