Let $X$ be a Markov chain, with countable state space $I$ and transition probability matrix $P$. $X$ is irreducible, but need not be recurrent. Let $S$ be a fixed subset of $I$.


Define a subset $K$ of $I$ to be "nice" if there exists $\epsilon = \epsilon\_K$ such that for all $k \in K$, $P\_{kS} \geq \epsilon$. (Here, $P\_{kS} = \sum\_{s \in S} P\_{ks}$.)


Given: with probability 1, there exists a nice set which $X$ visits infinitely often. (Note that the set $K$, and therefore the value of $\epsilon\_K$, may be random.)


Want to show: with probability 1, $X$ visits $S$ infinitely often.


It seems like it ought to be either trivially true or trivially false, but I'm failing to determine which...



Probability-one event for Markov chain
If I've understood your problem correctly, an argument along these lines
may help:




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Let ${\cal F}\_n=\sigma(X\_0,X\_1,\dots,X\_n)$ and define $S\_n=\left(X\_n\in S\right)$,
so that $S\_n\in {\cal F}\_n$.
We will use Levy's generalization of the Borel-Cantelli Lemma which states
that
$$\left( S\_n\mbox{ i.o.} \right)=\left(\sum\_n \mathbb{P}(S\_{n+1} | {\cal F}\_{n})=\infty\right).$$


Let's calculate the conditional probability.
Letting $E(x)=\{ X\_{n}=x\_{n},X\_{n-1}=x\_{n-1},\dots,X\_0=x\_0\}$ be a generic partition set, we get
\begin{eqnarray\*}
\mathbb{P}(S\_{n+1}\,|\,{\cal F}\_n)&=&\sum\_x\mathbb{P}(X\_{n+1}\in S\,|\,E(x))1\_{E(x)}\cr
 &=&\sum\_x\mathbb{P}(X\_{n+1}\in S\,|\,X\_n=x\_n)1\_{E(x)}\cr
 &=&\sum\_x P(x\_n, S)1\_{E(x)}\cr
 &=&P(X\_n, S),
\end{eqnarray\*}
where $P$ is the transition kernel for the Markov chain.


The definition of ``nice" set gives
$P(X\_n,S)\geq \varepsilon\_K 1\_K(X\_{n}),$
and since $(X\_n)$ visits $K$ infinitely often, we have 
$$\sum\_n P(X\_n,S)\geq \varepsilon\_K \sum\_n 1\_K(X\_{n})=\infty$$
almost surely.


