Proof
(related to Theorem: Trichotomy of Ordinals)
 Let \(\alpha, \beta\) be fixed ordinals and \(\gamma=\alpha\cap \beta\).
 By the lemma about equivalence of set inclusion and element inclusion of ordinals, it follows from \(\gamma\subseteq\alpha\) that \(\gamma\in\alpha\) or \(\gamma=\alpha\), and from \(\gamma\subseteq\beta\) that \(\gamma\in\beta\) or \(\gamma=\beta\).
 Altogether, we have the following cases:
 \(\gamma\in\alpha\) and \(\gamma=\beta\). From this case it follows that \(\beta\in\alpha\).
 \(\gamma\in\beta\) and \(\gamma=\alpha\). From this case it follows that \(\alpha\in\beta\).
 \(\gamma=\beta\) and \(\gamma=\alpha\). From this case it follows that \(\alpha=\beta\), and by the axiom of foundation, neither \(\alpha\in \beta\) nor \(\beta\in \alpha\) is possible.
 \(\gamma\in\beta\) and \(\gamma\in\alpha\). However, this case is not possible, since otherwise we would have \(\gamma\in\alpha\cap\beta=\gamma\), contradicting the axiom of foundation.
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References
Bibliography
 Hoffmann, Dirk W.: "Grenzen der Mathematik  Eine Reise durch die Kerngebiete der mathematischen Logik", Spektrum Akademischer Verlag, 2011
 Hoffmann, D.: "Forcing, Eine EinfÃ¼hrung in die Mathematik der UnabhÃ¤ngigkeitsbeweise", Hoffmann, D., 2018