I am trying to prove this statement, "if $P(S_n) \to 1$ as $n \to \infty$, prove that there exists subsequence $\{n_k\}$ such that $P(\cap_{n_k}S_{n_k}) > 0$".
As $lim_{n \to \infty} P(S_n) = 1 \ne 0 \Rightarrow \sum_{n=1}^{\infty} P(S_n) = \infty$.
If sequence $\{S_n\}$ is independent, then I can use Borel 0-1 law to get $P(S_n \text{ i.o}) = 1 \Rightarrow P(\{s : s \in S_{n_k} \}) = 1$ for k = 1,2, ... $\Rightarrow P(\cap_{n_k} S_{n_k}) = 1$.
On the other hand, how do I prove this statement without the independence condition ? Because I can not use Borel 0-1 law otherwise.
Just use a union bound.
For each $k\geq 1$ there exists $n_k$ with $\mathbb P(S_{n_k})>1-2^{-k}$, even if we also require $n_k>n_{k-1}$ for $k>1.$
That is, $\mathbb P\left(\overline{S_{n_k}}\right)<2^{-k}$, where the line denotes the complement event. So $$\mathbb P\left(\bigcup_{k\geq 1}\overline{S_{n_k}}\right)<\sum_{k\geq 1} 2^{-k}=1,$$ which means $$\mathbb P\left(\bigcap_{k\geq 1}S_{n_k}\right)>0$$ as required.