Wikipedia states that
Every well-ordered set (S,<) is order isomorphic to the set of ordinals less than one specific ordinal number [the order type of (S,<)] under their natural ordering.
I've been trying to find proofs on the Internet of this theorem at an appropriate level of rigor, but am entirely at a loss.
The best motivation I've found for defining von Neumann ordinals as the set of all lesser ordinals is that every well-ordering is order isomorphic to the set of all lesser ordinals under the ordinal comparison operation. Since I'm trying to use this theorem as a way to motivate defining ordinals, all I'd like to use in the proof is a supposition that ordinals are canonical representatives of order types, and not worry about how they're concretely defined.
I'm supposing that I've already proven that ordinal comparison is a well-ordering. How might I go about this proof? (Can I even do it without the von Neumann definition?) Thanks!
Clarification: My goal is not to show that every well-ordering is isomorphic to an ordinal. It is to show that every well-ordering is already isomorphic to a set of canonical representations of all lesser order types under ordinal comparison, so we might as well define ordinals as sets of all lesser ordinals, which is the von Neumann construction.
Best Answer
We start by a simple observation which is true for any totally ordered set, not only for well-ordered sets.
Let $(X,\le)$ be a linearly ordered set. We will use the notation $X_a=\{x\in X; x<a\}$ pre $a\in X$. Notice that these sets are initial segments of $X$.
Observation. Let $(X,\le)$ be a linearly ordered set and let $X'=\{X_a; a\in X\}$. Then the function $f\colon X\to{X'}$ defined as $$f(a)=X_a$$ is an isomorphism between the ordered sets $(X,\le)$ and $(X',\subseteq)$.
Proof. We can see immediately that $f$ is surjective. If $a<b$ then $a\in X_b$ and $a\notin X_a$, so $f(a)=X_a\ne X_b=f(b)$. The case $b<a$ is symmetric. So it is also injective.
The map $f$ is monotone: If $a \le b$ then $f(a)=\{x\in X; x<a\} \subseteq \{x\in X; x<b\}=f(b)$. (It suffices to notice that $x<a$ and $a\le b$ implies $x<b$ by transitivity.)
To see that also $f^{-1}$ is monotone, we notice that $X_a\subseteq X_b$ implies $a\le b$. (If we had $a>b$, then $b\in X_a\setminus X_b$, which contradicts $X_a\subseteq X_b$.) $\hspace{1cm}\square$
Side remark. Notice that we have used linearity of $\le$ in the above proof. The version of the above observation with $\{x\in X; x\le a\}$ instead of $X_a$ is true also for partially ordered sets and it implies that every partially ordered set is isomorphic to some system of sets ordered by inclusion (in fact, to a subset of $(\mathcal P(X),\subseteq)$.) See, for example, Theorem 1.11 in S. Roman: Lattices and Ordered Sets or this question.
How does this relate to ordinals? We can view ordinals as representatives or order types of well-ordered sets. (At least if we approach them in the naive rather than axiomatic way.) We can define inequality between ordinals via initial embeddings of the corresponding well-ordered sets. (I.e., ordinal type of $(A,\le)$ is less or equal to ordinal type of $(B,\preceq)$ iff $A$ is isomorphic to an initial segment of $B$. This seems to be the natural way how to compare ordinals/well-ordered sets.)
So the above lemma shows that a well-ordered set $X$ is isomorphic to the set of proper initial segments of $X$. Ordinal types of these initial segments are precisely the ordinals lesser than the ordinal type of $X$.