Let x be the least element in S that is not in z, and let y be the least element of T that is not in z. Suppose w ∈ x, yet w ∉ z. (Since x does not contain itself, w ≠ x.) Since S is transitive, w is a member of S. This would make w the least element of S that is not in z. This is a contradiction, hence z contains w after all. All the members of x belong to z. Conversely, if z contains w and w is not in x then x is in w, and since z is transitive, x is in z, which is impossible. Therefore all the members of z belong to x. Since x and z are subsets of each other, x = z. Similarly, y = z, hence x = y. Since x = y, it should be included in z. If z is truly complete then either x or y does not exist, whence z is all of S or all of T. The intersection of any two ordinals is one of the ordinals.
Given any two ordinals, one is a subset of the other. Subset is already a partial ordering, hence the ordinals are linearly ordered by subset.
Continuing the above, let S and T be ordinals with S∩T = T. If S is not equal to T, then T is a proper subset of S. Note that x, the least element of S outside of T, contains precisely the elements of T. Thus T = x and T ∈ S. When it comes to ordinals, T∈S iff T⊂S.
The class of ordinals is linearly ordered by membership, or by subset, as you prefer. We will now show the ordinals are well ordered. Keep in mind, this ordering is a formula, not a set. The symbols ∈ and ⊂ are expressions in logic that cannot be implemented as sets, since the domain, the class of all ordinals, is not a set. (We'll prove that later.)
Let S be a nonempty set of ordinals. Let x be any ordinal in S. Since ordinals are linearly ordered, every y in S is either in x, or contains x. Let T be the set of ordinals in S that are also in x. If T is empty then x is the least ordinal in S. Let's assume T is nonempty. The elements of x are well ordered, so let z be the least element of T. Now z is the least element in all of S, and the ordinals are well ordered, by set membership or by subset.
We can now show that the class of ordinals is not a set. Suppose S is the set of all ordinals. If y ∈ x ∈ S then x is an ordinal, and every member of an ordinal is an ordinal, so y is an ordinal and y is in S. Thus S is transitive. And we just showed the ordinals are well ordered by membership. If S does not contain itself then it is an ordinal, and S should contain S. Conversely, if S contains itself then it is an ordinal, and cannot contain itself. Therefore S cannot exist.