Let $I$ be a non-empty set and $(A_i)_{i\in I}$ a family of sets.
Is it true that there exists a subset $J\subset I$ such that $\bigcap_{j\in J}A_j=\bigcap_{i\in I}A_i$ and, for any $j_0\in J$, $\bigcap_{j\in J-\{j_0\}}A_j\neq\bigcap_{j\in J}A_j$?
If $I=\mathbb{N}$, the answer is yes (if I am not mistaken): $J$ can be constructed by starting with $\mathbb{N}$ and, at the $n$-th step, removing $n$ if that does not affect the intersection.
What if $I$ is uncountable? I guess the answer is still "yes" and tried to prove it by generalizing the above approach using transfinite induction, but I failed.
The answer "yes" or "no" and a sketch of a proof (resp. a counterexample) would be nice.