Notice: in this problem I shouldn't use any properties of the ordinals, and order types are not defined as ordinals.
Let $A,B$ be well ordered sets. We say that $otp(A) = otp(B)$, if $A,B$ are isomorphic, $otp(A) < otp(B)$ if $A$ is isomorphic to a proper initial segment of $B$, and $otp(B) < otp(A)$, the same.
I need to prove that the collection of all order types ($otp$'s) is well ordered.
For total ordering - anti-reflexivity is due to the fact that no set is isomorphic to any of its initial segments. Transitivity is due to the fact that if $otp(A) < otp(B), otp(B)<otp(C)$, it follows that $A$ is isomorphic to an initial segment of $B$ with $f:A\to B$, and $B$ is isomorphic to an initial segment of $C$ with $g:B\to C$. It follows that $range(f)$ is isomorphic to an initial segment of $C$, and therefore $g\circ f$ is an isomorphism between $A$ and an initial segment of $C$, i.e $otp(A) < otp(C)$.
Totality is due to the fact that every pair of well ordered sets are either isomorphic or one is embedded to the other.
I encountered a problem when trying to prove that every subset has a least element. I thought maybe showing, somehow, that there is a single initial segment that can be embedded into any other set in a set of well-ordered sets.