Is there a way to prove that $x > \sin^2(x) , x >0$

If $f(x)=\sin^2(x)$, then $f'(x)=2\sin(x)\cos(x)=\sin(2x)<1$, unless $x$ is of the form $\frac\pi4+k\pi$ ($k\in\mathbb Z$). So,$$x=\int_0^x1\,\mathrm dt>\int_0^xf'(t)\,\mathrm dt=f(x)=\sin^2(x).$$

Using $|\sin x | \leqslant 1 $ and the mean value theorem, for $x > 0$

$$\sin^2 x \leqslant |\sin x| = x\left|\frac{\sin x - \sin 0}{x} \right| = x|\cos \xi| \leqslant x$$