Every active finite route contains an active path

Among active routes with these endpoints choose one of minimum length. If a repeated vertex occurs at an endpoint, discard the initial segment through the later occurrence of $x$, or the terminal segment from the earlier occurrence of $y$. Endpoints are outside $C$, and all retained internal occurrences have their old incidences, so the shortened route is active. Consequently any repetition in a minimum route consists of two internal occurrences of a vertex $v$. Delete the segment between two such occurrences and identify them. Only the two incidences at the joined occurrence of $v$ change. If $v\in C$, its old occurrences were both colliders, so the joined occurrence is still a collider. If $v\mathrm{\notin }C$ and the joined occurrence is a noncollider, it is allowed; if it is a collider and $v$ is an ancestor of $C$, it is also allowed.

The sole remaining possibility is that the join makes $v$ a collider although $v$ is not an ancestor of $C$. Both original occurrences were then noncolliders. The retained edge before the first occurrence points into $v$, so its deleted departure edge must point out of $v$; likewise the deleted arrival edge at the second occurrence points out of $v$, since its retained departure edge points into $v$. Traverse the deleted segment starting with its arrow out of $v$. There must be a first place where its arrows cease to point forward: the final arrow points backward, and an entirely forward return to $v$ would also contradict acyclicity. At that first change is a collider $w$, reached from $v$ by a directed segment. A directed segment in a DAG has no repetitions. Activity of the original route makes $w$ an ancestor of $C$, hence $v$ an ancestor of $C$, a contradiction. Thus every loop deletion above is active and strictly shorter, contradicting minimality. The minimum route is therefore a path.