Let $X$ be a totally ordered set which has the least upper bound property and has an at most countable dense subset. Assume that $X$ has more than one element. Is $X$ isomorphic to some interval of $\mathbb R$?
Clarifications of terminology:
$Q\subseteq X$ is dense in a totally ordered set $X$ if and only if for all $x,y\in X$ such that $x<y$, there exists $q\in Q$ such that $x<q<y$.
Totally ordered sets $X$ and $Y$ are isomorphic if and only if there exists a monotonically increasing bijection $X\to Y$.
Intervals of $\mathbb R$ are $(a,b)$, $(a,b]$, $[a,b)$, $[a,b]$, $(a,\infty)$, $(-\infty,b)$, $[a,\infty)$, $(-\infty,b]$, and $\mathbb R$.
I know that the conclusion is not true when I just assume that $X$ is a linear continuum with more than one element. But I do not know what happens with my stronger assumption.