Generalized Second Borel-Cantelli lemma

The set equality in the lemma you mention only holds a.s., so it's not possible to prove that $\cap_{n \ge 1} A_n^C = \emptyset$. The best you can prove is that it has probability $0$. Apart from that, your proof looks correct.

To fix the proof, you just need to conclude from the lemma that since, as you correctly point out, $$\bigcap_{n\ge 1} A_n^C\subseteq \left\{\sum_{n \ge 1} P (A_n |\mathcal F_{n−1}) =\infty \right\},$$ then, up to a set of probability $0$, $\{A_n \ i.o.\}$ contains the set $\cap_{n\ge 1} A_n^C$. As these two sets are disjoint, this implies that $P(\cap_{n\ge 1} A_n^C)=0$.