Alternative proof Hall's marriage explanation

Yes, the proof is by induction on (let's say) $n$.

Hence, we may assume that for each $y \in A_n$ there is a tight $I \subset \{1,\dots,n-1\}$ with $y \in A_I$ (using induction).

Suppose that some $y \in A_n$ is not contained in any tight set. You should verify that this implies that $A_1 \setminus \{y\},\ldots,A_{n-1} \setminus \{y\}$ satisfies Hall's condition. By induction, there is a transversal, which you can extend to a transversal of $A_1,\ldots,A_n$ by adding $y$.

This type of proof is common in certain areas of graph theory: we show that if there is no tight set, then we can reduce the problem to a smaller one in one way, and if there is a tight set, then we can "cut along the tight set" to reduce the problem in another way.