Proving that the class of all ordinals is well ordered under $\in$
Solution 1:
For every $\gamma \in S$, we know that $\beta \subseteq \gamma$, so $\beta \in \gamma$ or $\beta = \gamma$. If $\beta \in \gamma$ for every $\gamma \in S$, then $\beta \in \bigcap S = \beta$, contradiction. Therefore, $\beta = \gamma$ for some $\gamma \in S$. Therefore, $\beta \in S$.
Solution 2:
For every $\gamma\in S$ we have $\beta\in\gamma\vee\beta=\gamma$.
So if $\beta\notin S$ then $\beta\in\gamma$ for every $\gamma\in S$ and consequently $\beta\in\bigcap S=\beta$.
A contradiction.