I'm just learning ordinal and cardinal arithmetic for the first time, and after learning that $n+\omega=\omega$ for finite $n$, I started to wonder whether $\omega+\omega_1=\omega_1$. This seems plausible both intuitively and by analogy with the previous fact. But I'm not sure how to prove it, and I can't find any theorems about ordinal arithmetic that let me prove it easily.
I know that $\omega_1$ is a limit ordinal, hence
$$\omega+\omega_1=\sup_{\alpha<\omega_1}\omega+\alpha.$$
Perhaps there is a theorem that says the sum of two ordinals of cardinality $\alpha$ must be another ordinal of cardinality $\alpha$?
Perhaps say that: $$ \omega + \alpha \ge \alpha\qquad \text{for all }\alpha < \omega_1 $$ therefore $$ \sup_\alpha (\omega+\alpha) \ge \sup_\alpha \alpha = \omega_1 $$ On the other hand, if $\alpha < \omega_1$ then $\alpha$ is countable, so $\omega+\alpha$ is countable. Thus $$ \omega+\alpha < \omega_1\qquad \text{for all }\alpha < \omega_1 $$ therefore $$ \sup_\alpha (\omega+\alpha) \le \omega_1 $$
You are right that we need: the sum of two countable ordinals is countable. This will follow from: the union of two countable sets is countable. (Maybe: depending on your definition of "ordinal".)
Another comment. This type or argument will not work for a general "limit ordinal" in place of $\omega_1$.