Hartogs' Lemma (Set Theory)/Proof 2

Proof
Let $W$ be the set of all well-orderings on subsets of $S$.

By Woset is Isomorphic to Unique Ordinal, there exists a mapping $F: W \to \operatorname{On}$ defined by letting $\map F s$ be the ordinal which is isomorphic to $s$.

By Mapping from Set to Ordinal Class is Bounded Above, $F \sqbrk W$ has an upper bound $\alpha_0$.

Then if $\alpha$ is any ordinal strictly greater than $\alpha_0$, then $\alpha \notin F \sqbrk {W}$.

Seeking a contradiction, suppose that there is an injection $g: \alpha \to S$.

Then by Injection to Image is Bijection, there is a bijection from $\alpha$ onto $g \sqbrk \alpha$.

But this induces a well-ordering on $g \sqbrk \alpha \subseteq S$ which is isomorphic to $\alpha$, contradicting the fact that $\alpha \notin F \sqbrk W$.