 Solution

Use two expansions of the rational order. In the first interpret $c_n=n$, so the named sequence is unbounded above. In the second interpret $c_n=-1/n$, so the named sequence is strictly increasing but has upper bounds in the rational order. Both expansions are countably infinite first-order models of the specified first-order theory.

An isomorphism preserving each $c_n$ would preserve the existence of an element greater than all the named constants, since any such element in one first-order structure would map to such an element in the other. The first first-order structure has none and the second has, for instance, $0$. They cannot be isomorphic. ThusThe property of lying above all $c_n$ is an infinite partial type rather than a single first-order sentence. This is why its realization can distinguish the two first-order models without contradicting the completeness established in (iv).