 Solution

Use the partial-isomorphism criterion for quantifier elimination: it is enough that every finite partial embedding between sufficiently saturated models have the one-point extension property.

Let $f:A\to B$ be a finite partial order isomorphism between models $M,N$ of the theory of dense linear order without endpoints, and take $c\in M\setminus A$. The finite set $A$ partitions $M$ into the points of $A$, the intervals between consecutive elements, and the two outer rays. The order relations between $c$ and $A$ identify one of those intervals or rays. The image $B$ determines the corresponding interval or ray in $N$. Density supplies a point there when it is bounded, and the absence of endpoints supplies one in either outer ray. Choose such a point $d$; then $f\cup \{(c,d)\}$ remains a partial order isomorphism.

The same argument works in the reverse direction. A back-and-forth method therefore extends finite partial order isomorphisms, making them partial elementary. The criterion proves quantifier elimination for dense linear orders without endpoints.