Proving the Kochen-Stone lemma using the Paley-Zygmund inequality
Here's a rough answer to your question. We have
$$ \frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n) -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2} {\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m) -\sum_{1\le m,n < \ell}\mathbb{P}\mathbb(A_n\cap A_m)}\geq \frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n) -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2} {\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)} $$
Fix $l \in \mathbb{N}_1$. Since $\lim_{k \rightarrow \infty}\sum_{n = 1}^k \mathbb{P}(A_n) = \infty$, by assumption, if $k$ is sufficiently large, $$ \left(\sum_{n=1}^{k}\mathbb{P}(A_n) -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2 \approx \left(\sum_{n=1}^{k}\mathbb{P}(A_n)\right)^2 $$
So if $k$ is sufficiently large, $$ \frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n) -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2} {\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)} \approx \frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)\right)^2} {\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)} = \frac{1}{\frac{\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)}{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)\right)^2}} $$
Suppose $$ \liminf_{k\to\infty}\frac{\sum_{1\le m,n \le k}\mathbb{P}(A_m\cap A_n)}{\left(\sum_{n=1}^k\mathbb{P}(A_n)\right)^2} = c < \infty $$
This means that no matter how large $k$ is, there is always some $k' \geq k$, such that $$ \frac{\sum_{1\le m,n \le k'}\mathbb{P}(A_m\cap A_n)}{\left(\sum_{n=1}^{k'}\mathbb{P}(A_n)\right)^2} \approx c $$
So for infinitely many $k$'s
$$ \mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right) \geq \frac{1}{\frac{\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)}{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)\right)^2}} \approx \frac{1}{c} $$
Since $\mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right)$ is decreasing in $k$, we have, approximately, $$ \lim_{k \rightarrow \infty}\mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right) \geq \frac{1}{c} $$
Since $l$ was arbitrary, $$ \lim_{l \rightarrow \infty}\lim_{k \rightarrow \infty}\mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right) \geq \frac{1}{c} > 0 $$