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Let $X$ be a linearly ordered topological space with a countable dense subset. Does it necessarily follow that $X$ is metrizable?

A linearly ordered space

EDIT: Apollo's comment int he answers implies the answer is a set negative. Let $X$ with be the open unit interval $(0,1)$ and adjoin to every real number $x$ a total order "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 open intervals form $(y, x]$ with $x$ and $y$ real and $[x',y)$ with $x'$ ghost and $y$ real form a basisfor $X$. For example, every ordinal can be given the order topology and the usual Euclidean topology on the these sets all contain a real line is the order topologyrational. See This space cannot be metrizable, because the Order subspace topology on Wikipedia. I am not asking for an example the set of a separable space that all ghost reals is not metrizable - there are many examples for exactly that (there's no need for of the Sorgenfrey line, you can even pick the trivial topology...).

Let $X$ be a linearly ordered topological space with a countable dense subset. Does it necessarily follow that $X$ is metrizable?
A linearly ordered space is a set $X$ with a total order such that the open intervals form a basis for $X$. For example, every ordinal can be given the order topology and the usual Euclidean topology on the real line is the order topology. See the Order topology on Wikipedia. I am not asking for an example of a separable space that is not metrizable - there are many examples for that (there's no need for the Sorgenfrey line, you can even pick the trivial topology...)
Let $X$ be a linearly ordered topological space with a countable dense subset. Does it necessarily follow that $X$ is metrizable?