Sine is 1-Lipschitz

If $a=b$ both sides are $0$. Otherwise assume $a<b$ (the other case is symmetric). The function $\sin$ is continuous on $[a,b]$ and differentiable on $(a,b)$, with derivative $\cos$.

By the Mean Value Theorem there is $c\in(a,b)$ such that
$$\sin b-\sin a=\cos c\,(b-a).$$

Since $|\cos c|\le1$, we get $|\sin b-\sin a|=|\cos c|\,|b-a|\le|b-a|$.