How to prove directly that the empty set is well ordered?

Apply the definition :

$\emptyset$ is well-ordered if it has a total order and every non-empty subset of $\emptyset$ has a least element in this ordering.

It means :

$\forall S (S \subseteq \emptyset \land S \ne \emptyset \to \ldots)$.

But the only subset of $\emptyset$ is $\emptyset$ itself.

Thus, $S \subseteq \emptyset \land S \ne \emptyset$ is False, and $(S \subseteq \emptyset \land S \ne \emptyset \to \ldots)$ is True, for every $S$.

You can prove it by exhaustion:

• for every pair of distinct elements of the empty set, the elements are comparable. This is vacuously true, so that you have a total order (whatever the comparison rule);

• for every non-empty subset of the empty set, there is a least element. This is vacuously true.