Odestheory12: Alternatively, take ordered sets $A=\{k\in \mathbb N: k\ne 3n \text{ for any } n\in \mathbb N\}$ and $B=\{3n: n\in \mathbb N\}$. Assuming the result that I wrote [earlier](https://chat.stackexchange.com/transcript/message/60153064#60153064) to be true, consider the sequences $(x_a), a\in \mathbb A$ and note that it converges to some L by the use of the linked result and the given condition. $(x_b)$ converges to L, is given.
So by the linked result, since $A\cup B=\mathbb N$ and that $A\cap B=\emptyset$, the sequence $(x_n)$ also converges.