I see this question being answered multiple times on this website including here but it still seems inconsistent with the mean value theorem.
If we have $\sin(x)$ defined on $[0,1]$ and we have $[x,y]\subset [0,1]$, then according to the mean value theorem, there exists $c\in (x,y)$ such that $|\sin(x)-\sin(y)|=\cos(c)|x-y|$.
Since $\cos(c)\in (0,1)$ for $c\in (0,1)$ doesn't this imply that $\sin(x)$ is indeed a contraction?
Everything you said in the question is correct, except for your conclusion that "$\sin(x)$ is indeed a contraction".
It is true (by MVT, like you say), that for every $0\leq x<y\leq 1,$ there is such a $c\in (0,1)$ such that $|\sin(x)-\sin(y)|=\cos(c)|x-y|$. But for $x$ and $y$ both approaching $0,$ this forces $\cos (c)$ to get very (arbitrarily) close to $1$ (from below $1$), and so there is no universal $c<1$ (or in wikipedia's notation $k<1$) that satisfies the contraction definition, as we require $c<1$ (or in wikipedia's notation $k<1$).
A simple way to think of a contraction is that for all $x\in D,\vert f'(x)\vert <t$ for some $t\in (0,1).$ That is, assuming $f$ is differentiable...