Intersection of Ordinals is Ordinal
Then $\bigcap A$ is an ordinal.
Let $i = \bigcap A$.
Let $n \in i$ and let $m \in n$.
Let $a \in A$.
By the definition of intersection, $n \in a$.
Thus $m \in a$.
As this holds for all $a \in A$, $m \in i$.
Thus $i$ is transitive.
Let $x,y \in i$.
Since $A$ is non-empty, it has an element $a$.
By the definition of intersection, $x, y \in a$.
Thus $x \in y$ or $y \in x$.
As this holds for all $x, y \in i$, $i$ is $\in$-connected.
Let $b$ be a non-empty subset of $i$.
Let $a \in A$ (such exists because $A$ is non-empty).
By Intersection is Largest Subset, $b \subseteq a$.
Thus $b$ has an element $x$ such that $x \cap b = \varnothing$.
As this holds for all such $b$, $i$ is well-founded.