 Solution

Use the completeness of the first-order theory of dense linear orders without endpoints and its quantifier elimination. After naming finitely many increasing constants, quantifier elimination still holds in the expanded first-order language: treat these constants as fixed parameters when eliminating quantifiers. Every quantifier-free sentence is a Boolean combination of logical equalities and order comparisons of the named constants, all decided by their stated strict increasing order. Hence ${\mathrm{DLO}}_n$ is complete.

An alternative formal proof of this finite case uses the stated general theorem that a first-order theory in a countable first-order language with only infinite first-order models and a unique countable first-order model is complete: if first-order models disagreed on a first-order sentence, downward Löwenheim-Skolem would give disagreeing countable first-order models. Part (iii) supplies the uniqueness.

Every first-order sentence of $L_{\omega }$ mentions only finitely many constant symbols, hence belongs to some $L_n$. Its truth is decided by ${\mathrm{DLO}}_n$, and every first-order model of ${\mathrm{DLO}}_{\omega }$ has a reduct satisfying that finite first-order theory. Thus the larger first-order theory also decides every first-order sentence. The rational-order examples in (ii) establish first-order consistency. ConsequentlyThere is no contradiction with (ii): completeness is agreement on individual finite first-order sentences, while categoricity requires a global isomorphism respecting all constants simultaneously.