One dimensional systems are often straightforward to analyse because solution are limited by equilibrium points (that is, there is no 'way around' an equilibrium point).
The system is globally Lipschitz hence a unique solution exists for all $\mathbb{R}$. It is also time invariant, so trajectories cannot cross.
Since $\sin$ is positive on $(0, \pi)$, we see that $\xi$ must be strictly increasing when $\xi(t) \in (0, \pi)$.
We have $\sin \pi = 0$, and $\zeta(t) = \pi$ is a solution to the system $\dot{x} = \sin x$ with $\zeta(0) = \pi$. Since the trajectories cannot cross, we must have $\xi(t) < \zeta(t) = \pi$ for all $ t$. Hence $\xi(t) \in (0,\pi)$ for all $t \ge 0$, and $\hat{\xi}=\lim_{t \to \infty} \xi(t) $ exists (and must be in $[\xi(0),\pi]$). We must have $\hat{\xi} = \pi$, otherwise we would have $\dot{\xi}(t) \ge \epsilon >0$ for some $\epsilon$, which would be a contradiction.
Hence $\hat{\xi}=\lim_{t \to \infty} \xi(t) = \pi$, and so the $\omega$-limit set is just $\{\pi\}$.
In general, you can see that if you start at $n \pi$, then the trajectory will remain there. If $\sin \xi(0) >0$, then the system will converge to $\pi \lceil {\xi(0) \over \pi } \rceil$, and similarly, if $\sin \xi(0) <0$, the system will converge to $\pi \lfloor {\xi(0) \over \pi } \rfloor$.
