This partial proof is taken from A Course in Mathematical Logic for Mathematicians by Yu. I. Manin.
Lemma. Let X and Y be two well-ordered sets. Then exactly one of the following alternatives holds:
(a) X and Y are isomorphic
(b) X is isomorphic to an initial segment in Y.
(c) Y is isomorphic to an initial segment in X.
PROOF.
Let X be well-ordered, and let $f:X \to X$ be a monotonic map. Then, for all $Z \in X$ we have $f(Z) \geq Z$. Therefore, $X$ is not isomorphic to any of its initial segments, $\hat X_1$. Now, let $X$ and $Y$ be well-ordered. We set $f = \{<X_1, Y_1>| X_1 \in X, Y_1 \in Y, \text{and there exists an isomorphism of } \hat X_1 \text{ with } \hat Y_1 \}$. First of all, $f$ is a graph of a one-to-one mapping of $\text{pr}_1f$ onto $\text{pr}_2 f$. In fact, if $X_1 \neq X_2$, say $X_1 < X_2$, then by the previously shown result, $\hat X_1$ is not isomorphic to $\hat X_2$...
I omitted parts of the proof that don't seem to be pertinent to the question (it's a bit lengthy). I'm confused by the notation $\text{pr}_1f$ and $\text{pr}_2f$. What does this mean? I don't think that it's defined anywhere else in the book*.
Also, this is unrelated, but can somebody provide me a link to an alternate proof (that's around the same level as this one) of this lemma? I like to see multiple proofs to make sure I really understand what's going on.
*The proof that I'm currently reading is in an appendix to a later chapter that the author recommends reading before continuing the book, so perhaps the notation is defined somewhere else in the book I just haven't gotten to yet (but I did look and didn't see anything)
Here $\operatorname{pr}$ is for projection: $\operatorname{pr}_1$ is the projection map to the first factor,
$$\operatorname{pr}_1:X\times Y:\langle x,y\rangle\mapsto x\,,$$
that takes each ordered pair in $X\times Y$ to its first component, and $\operatorname{pr}_2$ is the projection map to the second factor,
$$\operatorname{pr}_2:X\times Y:\langle x,y\rangle\mapsto y\,.$$
Note that here $f$ is a subset of $X\times Y$, so this does make sense. In more familiar terms, $\operatorname{pr}_1f$ is simply the domain of $f$, and $\operatorname{pr}_2f$ is the range of $f$.