$$ \newcommand{\Z}{{\mathbb Z}} \newcommand{\P}{{\mathbb P}} $$
Consider a biased simple random walk on $\Z$ with $P(k,k+1)=p,P(k,k-1)=q,P(k,k)=1-p-q$ with $p>q$. Let $\tau_k^+$ be the first return time to a state $k\in \mathbb{Z}$ i.e. $\tau_k^+=\inf\{t>\tau_k: X_t=k\}$ where $\tau_k=\inf\{t\geq 0:X_t=k\}$. Let $\P_i(X_t=x):=\P(X_t=x|X_0=i)$.
I'm trying to prove that $\P_0(\tau_0^+=\infty, \tau_1^+<\infty, \tau_3^+=\infty)>c>0$ for some constant $c>0$. In words, this event represents one where the walk starts at $0$, does not return to $0$ ever again, hits and returns to $1$, hits and does not return to $3$ ever again.
The fact that I need to consider trajectory of the random walk for infinite time is confusing me. Any suggestions?
The only way for a sequence to lie in the event $$\{\tau_0^+=\infty, \tau_1^+<\infty, \tau_3^+=\infty\}$$ is if the sequence goes $0 \to 1 \to S_{1,2} \to 1 \to 2\to S_{2} \to 3 \to 4 \to S_{\geq 4}$ where $S_{1,2}$ is a [possibly empty] finite sequence of $1$s and $2$s, $S_2$ is a [possibly empty] finite sequence of $2$s, and $S_{\geq 4}$ is a sequence of only $4$ and greater. Also, for any sequence in that event, the $S_{1,2}$, $S_2$, and $S_{\geq 4}$ are uniquely defined.
Importantly, the probability of such a sequence is, by the Markov property, equal to $$\mathbb{P}_0(0\to 1\to S_{1,2} \to 1 \to 2 \to S_2 \to 3 \to 4 \to \cdots) \times \mathbb{P}_4(4\to S_{\geq 4})$$
We can actually compute these probabilities, but it suffices to note that $$\begin{align*} \mathbb{P}_0&(0\to 1\to S_{1,2} \to 1 \to 2 \to S_2\to 3\to 4\to \cdots) \\ &\geq \mathbb{P}_0(0\to 1\to 1\to 2\to 3\to 4\to \cdots) + \mathbb{P}_0(0\to 1\to 2\to 1\to 2\to 3\to 4\to \cdots) \\ &= p^4(1-p-q) + p^5q \\ &= p^4(1-p)(1-q) \neq 0 \end{align*}$$ and $$\mathbb{P}_4(4\to S_{\geq 4}) = 1-(2-2(p+(1-p-q))) = 1-2q\neq 0$$ where in this last probability, we've used the result quoted in your previous question.