The sum of two well-ordered subsets is well-ordered

Ramsey theory! Suppose $A + B$ is not well-ordered. Then there is a strictly decreasing sequence $a_1 + b_1 > a_2 + b_2 > \cdots$. Observe that for any $i < j$, either $a_i > a_j$ or $b_i > b_j$ (or both). Make a graph with vertex set $\mathbb{N}$ by putting an edge between $i$ and $j$ if $a_i > a_j$, for any $i < j$. By the countably infinite Ramsey theorem, there is either an infinite clique or an infinite anticlique, and hence either a strictly decreasing sequence in $A$ or a strictly decreasing sequence in $B$, contradiction.

A more general result:

Let $G$ be an abelian totally ordered group. Let $E \subseteq G^+$. Write $E^*$ for the semigroup generated by $E$. If $E$ is well-ordered, then $E^*$ is well-ordered.

The case here is $G = (\mathbb R, +, <)$ and $E$ is an appropriate translate of $A \cup B$.

Attributed to Graham Higman with a simplified proof by C. St. J. A. Nash-Williams (the "minimal bad sequence" argument).

Higman, Graham, Ordering by divisibility in abstract algebras, Proc. Lond. Math. Soc., III. Ser. 2, 326-336 (1952). ZBL0047.03402.

Nash-Williams, C. St. J. A., On well-quasi-ordering finite trees, Proc. Camb. Philos. Soc. 59, 833-835 (1963). ZBL0122.25001.

There's also a more elementary proof than the one given by Nik Weaver (although I also enjoy the use of the countably infinite Ramsey's theorem!). First prove that an ordered set is well-ordered if and only if every sequence in it has a non-decreasing subsequence. Then given a sequence $(u_n)_{n \in \mathbb{N}}$ in $A+B$, for each $n \in \mathbb{N}$, pick the least $a_n \in A$ such that there exists a $b_n \in B$, which is then unique, with $u_n=a_n+b_n$. Extract a non-decreasing subsequence from $a$, then from $b \circ \varphi$ where $\varphi: \mathbb{N} \rightarrow \mathbb{N}$ is the first "extraction map".