Let $S_1$ and $S_2$ be two maximally independent sets. Assume without loss of generality that $ |S_1| \leq |S_2|$.

Suppose, for contradiction, that $|S_1| < |S_2|$. Then there exists some element $ x \in S_2 \setminus S_1 $ such that $S_1 \cup {x} \in I $.

But this contradicts the assumption that $S_1$ is maximally independent. Therefore, our assumption must be false, and we conclude that $|S_1| = |S_2|$.