The question is as follows,
Check that sum operation is "continuous" in the second variable, that's: if $\beta=\lim_{\eta<\gamma}(\beta_\eta)$ then $\alpha+\beta =\lim_{\eta<\gamma}(\alpha+ \beta_\eta)$.
How can I prove this?
If I'm not wrong, the limit of a sequence is the supremum of the set of the range of the sequence but I'm confused with that. If I take the sequence $(\beta_\eta)_{\eta<\gamma}$ where $\beta_\eta=2$, for all $\eta\in \gamma$ then,
$$ \beta=\lim_{\eta<\gamma}(\beta_\eta)=\beta=\lim_{\eta<\gamma}(2)=\bigcup2=1 $$
Where I'm wrong? Because the proof is straightforward if we just say that $\beta$ is a limit ordinal.
You're a bit off.
The limit of an increasing sequence is its supremum. Just like where in the real numbers, $\lim\limits_{n\to\infty}\frac1n=0$, does not mean that $0=\sup\{\frac1n\mid n\in\Bbb N\}$.
Moreover, if $\beta_\eta=2$, then your limit is not $\sup 2$, but rather $\sup\{2\}$, which is in fact $2$, and not $1$.
Continuity is defined the same way as it is defined elsewhere when topology is given. But the order topology on the ordinals has the benefit that sequences cannot be decreasing too much. Namely, if $\alpha_\xi$ is a convergent sequence of ordinals, then either it contains a strictly increasing sequence with the same limit (which is then the supremum), or that it is eventually constant.
Finally, what is true for most "beginner definitions" (e.g. ordinal arithmetic) is that they are defined using non-decreasing sequences, which then mean that limits are equivalent to suprema.