A chain of well ordered sets defines a "unique" well order on the union of chains? Halmos in his Naïve Set Theory, states the following result in Section 17, Well Ordering:

If a collection $\mathscr C$ of well ordered sets is a chain with respect to continuation, and if $U$ is the union of the sets of $\mathscr C$, then there is a unique well ordering of $U$ such that $U$ is a continuation of each set (distinct from $U$ itself) in the collection $\mathscr C$.

I think that this is not quite correct. Consider this. $\mathscr C = \{ \{0, 2\}, \{0, 2, 3\}, \{0, 2, 3, 6, 8\} \}$ with the usual ordering of $\mathbb N$ in each of the sets. Now, I can order $U = \{0, 2, 3, 6, 8\}$ as $0 < 2 < 3 < 8 < 6$. Then $U$ is a continuation of each $X\in\mathscr C\setminus \{U\}$. But the usual order of $\mathbb N$ also satisfies this requirement, albeit being a different order.
Question: What has gone wrong here?

He defines continuation as follows. A well ordered set $A$ is a continuation of a well ordered set $B$ iff $B$ is an initial segment (the set of all strict predecessors) of some element of $A$ and if the order on $B$ can be inherited from that on $A$. (Hence, a well ordered set can't be a continuation of itself.)
 A: Halmos's statement is indeed incorrect as written.  It should instead say require the ordering on $U$ to be a nonstrict continuation of the ordering on every element of $\mathscr{C}$ (including possibly $U$ itself), where "nonstrict continuation" means being either an initial segment or the same set with the inherited order.  So, in the case that $U$ is an element of $\mathscr{C}$ already the result is trivial since you just have to use the given ordering on $U$.
(Essentially, what happened here is that Halmos tweaked his statement with the parenthetical "(distinct from $U$ itself)" because of the issue that his notion of "continuation" is strict and so $U$ cannot be a continuation of itself.  However, he missed that omitting $U$ from the requirement (rather than fixing the notion of "continuation" to allow it) breaks the uniqueness claim, because there might be some elements of $U$ that are not contained in any other elements of $\mathscr{C}$ and so if you don't require any compatibility with the ordering of $U$ you don't know how those elements must be ordered.)
