EDIT: This has to be done without calculus methods. I.e. noting $h'\ge 0$ won't do. I have used the mean value theorem, however this has been deemed permissible.
Prove $\sin(x) \le x$ for all $x\ge0$.
Consider the function $h(x)=x-\sin(x)$. Then, $h(0)=0-\sin(0)=0$
Consider an interval $[0,y]$ for $y>0$. $h(x)$ is continuous on $[0,y]$ and differentiable on $(0,y)$, thus the conditions of the mean value theorem are satisfied. Therefore,
$$\exists c\in(0,y): h'(c)=\frac{h(y)-h(0)}{y-0}=\frac{h(y)}{y} $$
Note that $h'(x)=1-\cos(x)$. This means $0 \le h'(x) \le 2$, paticualrly $h'(c) \ge 0$.
Then, $$\frac{h(y)}{y}\ge0$$ We know $y>0$ so $$h(y) \ge0 \implies y-\sin(y)\ge0 \implies y\ge\sin(y) \text{ for } y\ge0 $$ as required. Is this proof valid?