Linear order where every initial segment is finite

Question

It's true that a set with a linear order where every initial segment is finite is necessarily countable? I don't think so but I can't find a counterexample (or make a proof)

Answer 1

Yes. You can even prove the following:

If $(A,\leq)$ is a linear order that every proper initial segment is finite, then $(A,\leq)$ embeds, as an ordered set, into an initial segment of the natural numbers.

I'll give you a hint: think about a natural way to associate each element with a natural number, and show that is a unique and order preserving way to do so.

This means that $A$ is either finite, or countable.