Let $X$ be a linearly ordered topological space with a countable dense subset. Does it necessarily follow that $X$ is metrizable?
EDIT: Apollo's comment int he answers implies the answer is negative. Let $X$ be the open unit interval $(0,1)$ and adjoin to every real number $x$ a "ghost number" $x'$ such that $x'$ is the immediate successor of $x$. The "real rationals" are dense in this space. Simply note that sets of the form $(y, x]$ with $x$ and $y$ real and $[x',y)$ with $x'$ ghost and $y$ real form a basis, and these sets all contain a real rational. This space cannot be metrizable, because the subspace topology on the set of all ghost reals is exactly that of the Sorgenfrey line.