This question requires using transfinite induction. I plan to fix $\alpha$ and then do transfinite induction on $\beta$.
Recall that an ordinal $\alpha$ is countable if $\alpha < \omega_1$, where $\omega_1$ is the first uncountable ordinal.
The successor of an ordinal $\alpha$ is the smallest ordinal number greater than $\alpha$. An ordinal number that is a successor is called a successor ordinal. An ordinal that is not a successor ordinal is either a limit ordinal or 0.
By the definition of ordinal addition, if $\beta$ is a successor ordinal then $\beta = \gamma + 1$ and so $\alpha + \beta = (\alpha + \gamma) + 1$. If $\beta$ is a limit ordinal, then $\alpha + \beta = \sup \{\alpha + \gamma \mid \gamma < \beta\}$.
Recall that transfinite induction is made up of three parts: Base case: Set $\beta = 0$. Successor case: Set $\beta = \gamma + 1$ for an ordinal $\gamma$. Limit case: Let $\beta$ be a limit ordinal.
I have managed to prove the base case (trivial) and the limit case, since this follows from the fact that the union of two countable sets is countable. However, I am stuck on proving the successor case.
Any help would be greatly appreciated and thanks in advance.