Here is the proof given by a textbook:
If $L\ne\infty$ and $\alpha>L$, the set $\{n:s_n>\alpha\}$ is finite; otherwise $\sup\{s_n:n>N\}>\alpha$ for all $N$ and hence $L=\lim\sup s_n\ge\alpha>L$, which is a contradiction.
However I am confused when it says "$\sup\{s_n:n>N\}>\alpha$ for all $N$ "$\implies$ "$L=\lim\sup s_n\ge\alpha$". Where is the connection?
Recall that $$ L=\limsup_{n\to\infty}s_n=\inf_{N\geq 1}\sup\{s_n:n>N\}$$
Therefore if you know that $t_N=\sup\{s_n:n>N\}>\alpha$ for all $N\geq 1$, then it follows that $\alpha$ is a lower bound for the set $\{t_N:N\geq 1\}$. As $L$ is the greatest lower bound of this set, this implies that $L\geq \alpha$.