real analysis - Showing $sin(x) 0$ using the mean value theorem

I want to show that $\sin(x) < x$ for all $x>0$, using the mean value theorem.

Since the sine is bounded above by $1$, it's obviously true for $x > 1$. Consider $x \in ]0,1]$. Let $f(x)=\sin(x)$. Choose $a=0$ and $x>0$, then there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with

$$f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow (\sin(x))'(x_0)= \frac{\sin(x)-\sin(a)}{x} \Leftrightarrow \cos(x_0)=\frac{\sin(x)}{x}$$

Since $1\geq x_0>0 \Rightarrow \cos(x_0) < 1$,

$$\Rightarrow 1 > \cos(x_0)=\frac{\sin(x)}{x} \Rightarrow x > \sin(x)$$

Is my proof correct?