When we say that a totally ordered set $X$ has the least upper bound property, we mean that every upper bounded subset of $X$ has a least upper bound.
I know that a totally ordered set with the least upper bound property is locally compact (with respect to the order topology). Is the reverse also true?
No. Consider $X=(0,1)\cup (2,3)$ with the standard ordering. It is of course locally compact, but it doesn't have the least upper bound property: every upper bound of $(0,1)$ has to be inside $(2,3)$, and so a smaller one exists.