Linear order where every initial segment is finite
Solution 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.