Ordinal Subset is Well-Ordered

Theorem
Let $S$ be a class.

Let every element of $S$ be an ordinal.

Then $\left({S, \in}\right)$ is a strict well-ordering.

Proof
Let $\operatorname{On}$ denote the class of ordinals.

By definition of subset, $S \subseteq \operatorname{On}$.

But $\operatorname{On}$ is an ordinal.

Therefore $\operatorname{On}$ is well-ordered by $\in$.

This means that $S$ is also well-ordered by $\in$.