Consider Markov chain $\{X_n\}_{n=0}^\infty$ on a finite state space $V = [k] = \{1,...,k\}.$ Let $\mathcal{A}$ be the set of absorbing states ($i \in \mathcal{A}$ $\Rightarrow$ $p_{i,i} = P(X_1 = i |X_0=i) = 1$). Assume that for all $i \notin \mathcal{A}$ and all $j \in V$ there is $n \geq 1$ such that $P(X_n=j|X_0=i) > 0$, then $P(X_n \in \mathcal{A} \text{ for all large n} |X_0=i) = 1$ for all $i \in V$.
How to start? What we know is that starting at any state (absorbing or not) we can get to an absorbing state and never leave once we are there. If $i \in \mathcal{A}$ we are done since once we are in here we never leave. If $i \notin \mathcal{A}$, then there $n \geq 1$ such that $P(X_n \in A|X_0=i) > 0$ and once we are in here then again we do not leave. But we are only given that this probability is positive, not necessary guaranteed to happen. Also this is not rigorous at all, how could I write this more rigorous?
Edit: I stated the question incorrently but it is fixed now.
Assume that for all $i\in V$ there is $n_i \geq 1$ and $a_i \in \mathcal A$ such that $P(X_{n_i}=a_i|X_0=i) > 0$. Write $$\delta:= \min_{i \in V} P(X_{n_i}=a_i|X_0=i) > 0 $$ and $n=\max_{i \in V} n_i$. Then by induction on $\ell$, we can check that for all $i\in V$, $$P(X_{\ell n} \notin A |X_0=i) \le (1-\delta)^\ell\,.$$ Indeed, after each round of $n$ steps, the probability of absorption is at least $\delta$, i.e., the probability the chain is not absorbed, given that it has not been absorbed so far, is at most $(1-\delta)$, so after $\ell$ rounds the probability the chain is not absorbed is at most $(1-\delta)^\ell$.
It follows from the Borel-Cantelli Lemma (or directly) that $$P(X_{m} \notin A \; \text{for infinitely many } \; m|X_0=i) =0 \,,$$ which is equivalent to the requirement of the problem.