Let $G = (V,E)$ be an graph which is locally finite (every vertex has only finitely many edges - but there may be infinitely many vertices), and connected. Let $X_n$ be a random walk on $G$ that follows the transition probabilities induced by the edge weights of $G$.
Is it possible to find a $G$ such that there exists a nonempty $U \subseteq V$ such that if we let $W$ be the subgraph of $G$ induced by the vertices $V \setminus U$, then $W$ is connected and we have that for any $w \in W$, $\lim_{n \rightarrow \infty}P_w(X_n \in W | X_{n-1} \in W) = 1$ ?
($P_w$ denotes that $X_0 = w$)
Edit: Made some mistakes earlier. Corrected to indicate that the walk starts in $W$ and that both $W$ and $G$ are connected.
One way to get $$ \lim_{n \rightarrow \infty}P_w(X_n \in W | X_{n-1} \in W) = 1 $$ is to take a graph whose random walk Markov chain is transient.
For instance, let $G$ be an infinite binary tree, and let $W$ be one of the branches from the root. Then with probability $1$ the random walk only returns to the root a finite number of times, so in the limit as $n \to \infty$ the probability of the random walk leaving $W$ is $0$.