Intersection of Ordinals is Smallest
Then $m$ is the smallest element of $A$.
By Intersection is Largest Subset, $m \subseteq a$ for each $a \in A$.
It remains to show that $m \in A$.
Let $m^+ = m \cup \set m$ be the successor of $m$.
- for each $a \in A$, either $m^+ \subseteq a$ or $a \in m^+$.
Let $m^+ \subseteq a$ for all $a \in A$.
Then by Intersection is Largest Subset $m^+ \subseteq m$.
Thus there is an $a \in A$ such that $a \in m^+$.
Thus $a \in m$ or $a = m$.
Let $a \in m$.
Then $a \subsetneqq m$.
This contradicts the fact that $m \subseteq a$.
Thus $a = m$.
Thus it follows that $m \in A$.