how to strictly prove $\sin x<x$ for $0<x<\frac{\pi}{2}$

Define the function $f(x)=x-\sin x.$ Observe that $f(0)=0$ and $f'(x)=1-\cos x \geq 0$. The derivative is equal to $0$ only at isolated points, so the function increases in the interval $[0, \infty)$. That is, for all $x>0$ we have $f(x)>f(0)=0$. Thus $x>\sin x$ for all $x>0$.

Define the function $f(x)$ as

$$f(x) = \int_0^x\frac{1}{\sqrt{1 - t^2}} \,dt \tag 1$$

for $|x|\le 1$, as the inverse function for the sine function. That is to say that $\sin(x)=f^{(-1)}(x)$.

Clearly, $f(x)$ as given in $(1)$ is monotonically increasing with $0=f(0)\le f(x)\le f(1)=\pi/2$. Furthermore, it is easy to see from $(1)$ that $f(x)\ge x$ for $x\ge 0$ with $f(x)>x$ for $0<x\le 1$.

Inasmuch as $f$ is monotonically increasing on $(0,1)$, its inverse, $\sin(x)$, is monotonically increasing with $0\le \sin(x)\le 1$ for $0\le x\le \pi/2$. Moreover, we see that since $f(x)>x$ for $0<x\le 1$, then $0< \sin(x) <x$ for $0 <x\le \pi/2$.

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EVALUTING THE LIMIT $\displaystyle \lim_{x\to 0}\frac{\sin(x)}{x}$

Using standard analysis (e.g. L'Hospital's rule), and without appeal to any a priori knowledge of properties of $\sin (x)$ and $\cos (x)$, one can derive easily the limit sought.

(1) $f(0) = 0$ implies $\sin (0) = 0$ since $f(x)$ is the inverse function of $\sin (x)$;

(2) $\frac{df(x)}{dx} = \frac{1}{\sqrt{1 - x^2}}$ follows from the fundamental theorem of calculus;

(3) $\frac{d\sin (x)}{dx} = \sqrt{1 - \sin^2 (x)}$ using (2) along with the relationship between derivatives of inverse functions;

(4) $\lim_{x\to 0} \frac{\sin (x)}{x} = \lim_{x\to 0} \frac{d \sin (x)}{dx}$ follows from L'Hospital's Rule;

(5) Using (1)-(4), $\lim_{x\to 0} \frac{\sin (x)}{x} =\lim_{x\to 0} \frac{d \sin (x)}{dx} = \lim_{x\to 0} \sqrt {1 - \sin^2 x} = 1$

which completes the proof.

And this proof did not appeal to any prior knowledge of the sine function and did not even mention the cosine function!

This is probably the simplest, rigorous proof available.

We can define $\sin x$ as power series. Applying the knowledge of power series, obtain the derivative of $\sin x$, and then we will easy prove the inequality. Concluding geometry of $\sin x$, please refer to this.