Ordinals are Well-Ordered/Proof 2

Proof
By Ordinals are Totally Ordered, the ordinals are totally ordered by $\subseteq$:

Thus:
 * the strict ordering $\subsetneqq$ on ordinals

and
 * the strict ordering $\in$ on ordinals

are the same.

Suppose the ordinals were not well-ordered by $\subsetneqq$.

Then we could find a sequence $\left \langle{X_n}\right \rangle_{n \mathop = 0}^\infty$ of ordinals such that:
 * $X_0 \supsetneqq X_1 \supsetneqq X_2 \cdots$

So for all $n > 0$:
 * $X_n \subsetneqq X_0$

so:
 * $X_n \in X_0$

Thus $\left \langle{X_{n+1}}\right \rangle_{n \mathop = 0}^\infty$ is a decreasing sequence under $\subsetneqq$ of elements of $\left \langle{X_n}\right \rangle$.

But since $X_0$ is an ordinal it is well-ordered by $\subsetneqq$.

From Condition for Well-Foundedness, this demonstrates a contradiction.