TITLE: How to prove directly that the empty set is well ordered?
QUESTION [5 upvotes]: I found a post concerning this question, but I cannot understand the proof given there of the fact I'm talking about. 
The link is : Verification of proof that the empty set is well ordered
I think I can prove indirectly that the empty set is well ordered in the following way : 
(1) suppose the empty set is not well ordered
(2) that is, suppose it is false that every non empty subset of the empty set has a first element 
(3) it means there exists at least some set S such that 

S is a non empty subset of the empty set 
and S has no first element

(4) which requires the first conjunct  " S is a non empty subset of the empty set" to be true
(5) but this is impossible, for the empty set has only one subset, which is empty. 
However , I cannot manage to give a direct proof of the same fact. 
I cannot go further than this : 
(1) Let S be an arbitrary set 
(2) Assume it is true that :  S is a non empty subset of the empty set 
(3) Derive from this that : S has a first element. ... but how?

REPLY [6 votes]: 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 $S$.

REPLY [4 votes]: 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.