Is the union of finite and infinite sequences over alphabet of length 1 countable?
You haven't defined what you mean by "finite or countably infinite sequence" over an alphabet $\Sigma$, so let me assume that a "finite sequence" is a word over $\Sigma$, while a "countably infinite sequence" is an $\omega$-word over $\Sigma$, that is, a mapping from $\omega$ (the set of natural numbers) to $\Sigma$.

In your case, there is exactly one word of each length $0,1,2,\ldots,\omega$, so the set of "finite and countably infinite sequences" is countable, since the set $\omega + 1 = \omega \cup \{\omega\}$ of possible lengths (which is in one-to-one correspondence with the possible "finite and countable infinite sequences") is countable.

If you want to be very formal, you can show that $\omega + 1$ is countable by giving an explicit bijection between $\omega$ and $\omega + 1$, such as
$$
0 \mapsto \omega, \quad n+1 \mapsto n \text{ for all } n \geq 0.
$$