 Solution

Because the vertex set is finite, at least one vertex occurs infinitely often, so $\mathrm{Inf}(\rho )$ is nonempty. Let $u,v$ belong to this set. After any chosen visit to $u$, a later visit to $v$ exists; the intervening edges give a directed walk from $u$ to $v$. After that visit to $v$, a later visit to $u$ also exists, giving a walk in the other direction. Removing repeated portions gives directed paths if desired.

Thus every pair in $\mathrm{Inf}(\rho )$ is mutually reachable and belongs to the same strongly connected component. This proves the claimed inclusion, which need not be equality: an infinite walk can repeatedly follow one cycle while omitting other vertices in its component. This is the recurrence-set part of the infinite walk criterion in a finite directed graph.