Call a rule $\Phi:\Bbb{ON}\rightarrow\Bbb{ON}$ normal if for any ordinal $\alpha$, $\Phi(\alpha)<\Phi(\alpha^+)$, and for limit ordinals $\lambda$, $\Phi(\lambda)=\bigcup_{\alpha<\lambda}\Phi(\alpha)$.
Prove that for all non-zero ordinals $\beta$, the rule $\alpha\mapsto\alpha+\beta$ is not normal. (and that $\alpha\mapsto\beta+\alpha$ is normal, but that's an easy induction)
Now, possibly the question is ill-formulated and wants us to prove only that that rule is not normal for some $\beta$. I have the solution below, but I don't see why it provides an answer. To me, it proves that for $\alpha<\beta .\omega$ we have $\alpha + \beta.\omega<\beta.\omega + \beta.\omega$, but that doesn't seem to go against the normality condition for limit ordinals...
Am I not seeing something obvious?
Here's an alternative (easier, I think) proof: Let $\Phi(\alpha) = \alpha+\beta$.
If $\omega\leq \beta$, then $\Phi(\alpha+1) = \alpha+1+\beta = \alpha+\beta = \Phi(\alpha)$, so $\Phi$ fails to be strictly increasing.
If $0<\beta<\omega$, then $\Phi(\omega) = \omega +\beta$, while $\bigcup_{n<\omega}\Phi(n) = \bigcup_{n<\omega} (n+\beta) = \omega$, so $\Phi$ fails to be continuous.
I agree that the solution you posted doesn't answer the question: it seems to me that it shows that the function $\alpha\mapsto \alpha+\beta\cdot\omega$ fails to be continuous for all non-zero $\beta$, rather than $\alpha\mapsto \alpha+\beta$.
But the same idea can be easily adapted to answer the question. I'm sure this is what the person who wrote the solution originally intended.
Suppose $\alpha<\beta\cdot\omega$. Then $\alpha\leq \beta\cdot n$ for some $n\in \omega$, so $$\Phi(\alpha) = \alpha+\beta \leq \beta\cdot n + \beta = \beta\cdot (n+1) < \beta\cdot \omega.$$ It follows that $\bigcup_{\alpha<\beta\cdot \omega}\Phi(\alpha) = \beta\cdot \omega$. But $$\Phi(\beta\cdot\omega) = \beta\cdot\omega + \beta = \beta\cdot(\omega+1).$$ So $\Phi$ fails to be continuous.