A non-separable, connected ordered topological space with no interval order isomorphic to the reals

Question

Is there a non-separable, connected ordered topological space such that for each point $x$ in that space, there is no interval containing $x$ that is order isomorphic to real line?

Answer 1

I believe an example can be found using an Aronszajn continuum. A densely ordered connected linearly ordered space $X$ is an Aronszajn continuum if

1. $X$ is not separable,
2. $X$ is first-countable, and
3. $\overline A$ is second-countable for every countable $A \subseteq X$.

Such spaces are the Dedekind completion of a densely ordered Aronszajn line. In particular, these can be proven to exist.

Given an Aronszajn continuum $X$ with linear ordering $\leq$, we take the quotient by the equivalence relation $$a \sim b \leftrightarrow \{ x \in X : a \leq x \leq b \}\text{ is separable.}$$ The resulting space $X/\mathord{\sim}$ is also an Aronszajn continuum, but has the added property that no non-degenerate interval is separable, which implies that no point is contained in an interval homeomorphic/order-isomorphic to the reals.

---

_{Note: It's possible that any Aronszajn continuum will satisfy the requirements. An Aronszajn line has the property that no uncountable subset is order-isomorphic to a subset of $\mathbb R$, which might carry over to the Dedekind completion, but I haven't been able to figure out the details.}