 Solution

Let $d_t$ be the graph distance between fox and rabbit after round $t$, with $d_0$ their initial graph distance. Initial reachability makes $d_0$ finite, and ordinary moves preserve membership in their common connected component of a graph. After the rabbit's next move, the triangle inequality gives intermediate separation at most $d_t+1$. If that separation is at most two, the fox catches the rabbit that round. Otherwise the fox takes two steps along a shortest path, leavingThus a successful surviving history has at most $d_0-1$ rounds; if it does not escape by then, the next round captures it. The shortest-path pursuit distance bound excludes infinite live branches, even when the rabbit stays. Together with the finite branching factor, it makes the whole relevant search tree finite, so depth-first tree search eventually finds an escape whenever one exists.

For nonoptimality, take the corridor of part (g) and let depth-first tree search try left first. It immediately returns the one-round escape with bonus $1$, whereas trying right twice earns $2$. The decreasing separation proves termination, not optimality of the first returned escape.