 Solution

By choice-function well-ordering construction, use transfinite recursion to choosewhenever the set on the right is nonempty. The chosen elements are distinct, so if this construction continued through the Hartogs theorem ordinal $h(x)$, the map $\beta \mapsto x_{\beta }$ would inject $h(x)$ into $x$, a contradiction. It therefore stops at some ordinal $\alpha <h(x)$. At the stopping stage every element of $x$ has been selected, and henceis a bijection $\alpha \to x$.

Assuming the axiom of choice, every set $x$ has such a choice function on its nonempty subsets. Transporting the membership order on $\alpha$ across the bijection well-orders $x$. Thus the axiom of choice implies the well-ordering theorem.