Proposition: A dense distinguishing order without an interval basis

There is an interpolating order whose strict pasts and strict futures each distinguish points, and whose points all have predecessors and successors, but whose chronological intervals do not form a basis.

Proof. Use

Order each negative branch by its rational coordinate, put both branches below the nonnegative chain, and order that chain normally. The two negative branches are mutually incomparable. Let $\ll$ be strict order.

Rational density gives interpolation, including across the joining point by choosing a later negative point on the same branch. Pasts and futures distinguish points within a branch by density, and distinguish different branches through their negative elements. Every point has predecessors and successors.

For $a=(-1,A)$, $b=(-1,B)$, and $c=1$,

Every interval containing $0$ has a lower endpoint on one negative branch, so contains negative points of that branch. No such interval is contained in the displayed intersection.