Let $W \neq \emptyset$ be a set of well-ordered sets. Show that there is $A \in W$ that for all $B \in W$, $A$ is order-isomorphic to $B$ or to an initial segment of $B$.
This is from Goldrei, exercise 7.66. Being order-isomorphic to an initial segment is a strict linear order on well-ordered sets, hence the title.
If $\emptyset \in W$ then we can take $A=\emptyset$. In general we should take the set in $W$ with the least order type, but Goldrei asks it before he develops order types, knowing only that this order is linear. So is there any more straightforward solution?
Pick any $B\in W$. Consider all initial segments in $B$ that are order isomorphic to some $A\in W$. There is a smallest such, since $B$ is well ordered. Take the corresponding $A$.