I want to find an ordinal number $\beta$ such that $\sup\{\alpha_1,\alpha_2,\dotsc\}+1<\beta$ for any (countable) sequences $\alpha_1,\alpha_2,\dotsc< \beta$. I know that the least uncountable ordinal number $\omega_1$ is an example of $\beta$.
Does such $\beta$ less than $\omega_1$ exsit?
The first uncountable ordinal is the smallest ordinal to have this property. That's because for any countable ordinal $\beta$, you can actually enumerate $[0,\beta)$ as a sequence $\{\alpha_1, \alpha_2, \ldots\}$, precisely because $\beta$ is countable. Then $\sup\{\alpha_1+1,\alpha_2+1,\ldots\}=\beta$.
In a partially ordered set $(P,<)$, a subset $A$ of $P$ is said to be cofinal if for any $x\in P$, there exists $a\in A$ such that $x\leqslant a$. The cofinality of a set is the least cardinality of a cofinal subset. Each ordinal $\beta$ is (or can be) defined so that $\beta=[0,\beta)$. That is, each ordinal is the set of all smaller ordinals. So the cofinality of an ordinal $\beta$ is the cofinality of $[0,\beta)$. For $0$, this is zero, since the empty set is cofinal. The cofinality is $1$ if and only if $\beta$ is a successor. What you are asking about is ordinals with uncountable cofinality. Indeed, for a non-zero $\beta$, the existence of a sequence $(\alpha_n)_{n=1}^\infty \subset [0,\beta)$ with $\sup \alpha_n +1=\beta$ is equivalent to having cofinality which is at most countable.
Every countable ordinal has countable cofinality, because the entire set $\beta=[0,\beta)$ is cofinal in itself, and is countable.
There exist uncountable ordinals with countable cofinality (eg $\omega_1+1$, which has cofinality $1$, or $\omega_1\omega$, which has cofinaly $\aleph_0$, since $\sup_n \omega_1n=\omega_1\omega$.