Let $L$ be a linear order. if $A \subset L$ and $a \in L$, then $a$ is called a strict upper bound if $x <a$ for every $x \in A$. Now suppose the following is true for every $A \subseteq L$: if $A$ has a strict upper bound, then $A$ has a least strict upper bound. Prove that if $L \not= \emptyset$, then $L$ has a least element.
I think I have a counterexample against this, namely $\mathbb{Z}$. This set has a linear order and every finite subset can be bounded with a strict upper bound. Let $X \subsetneq \mathbb{Z}$ and let $X$ be finite. Then $X$ has a maximal element $k$, then $k+1$ is the least strict upper bound for $X$. However, $\mathbb{Z}$ obviously has no least element in its total order. So how can this theorem be true?
Let $A$ be the empty subset of $L$. Any element of $L$ is a strict upper bound for $A$ since our condition becomes empty. Since $L$ is empty, $A$ has a strict upper bound, hence a least strict upper bound, which is a least element.
In your example it goes wrong for the empty set (see comments above).