First, $\sin^{-1}(x)$ is defined for $-1 \le x \le 1$. Consequently, the proof implicitly presumes that $-1 \le x \le 1$.
Second, you did not write $\iff$ in each step.
Third, your proof did not specify why $\cos^2(\sin^{-1}(x))=1-x^2 \iff \cos(\sin^{-1}(x))=\color{red}{+}\sqrt{1-x^2}$. You need to justify that by mentioning that $-\pi/2 \le \sin^{-1}(x) \le \pi/2$ and $\cos(\theta) \ge 0$ for $\theta \in [-\pi/2,\pi/2]$.
A second of thinking tells you that no function achieves that !
$$\sqrt{x+y}\ne\sqrt x+\sqrt y,\frac1{x+y}\ne\frac1x+\frac1y,\log(x+y)\ne\log x+\log x,\cdots$$
$$\frac xy+1\ne\frac{x+1}{y+1},\tan\frac xy\ne\frac{\tan x}{\tan y}\cdots$$
There are just two exceptions:
$$a(x+y)=ax+ay$$
and
$$\left(\frac xy\right)^a=\frac{x^a}{y^a}.$$
So you'd better ask why the linear function is additive and why the power function is multiplicative.
If you want to find all additive functions, i.e. such that
$$f(x+y)=f(x)+f(y),$$ you immediately see that
$$f(2x)=2f(x)$$ and by induction
$$f(nx)=nf(x).$$
This generalizes to rationals,
$$qf\left(\frac pqx\right)=qpf\left(\frac xq\right)=pf(x)\implies f\left(\frac pqx\right)=\frac pqf(x),$$ and to reals
$$f(rx)=rf(x),$$ but the proof is more technical.
Now, setting $r\to x,x\to1$,
$$f(x)=f(1)\,x=ax.$$
For the multiplicative functions
$$g(xy)=g(x)g(y)$$
consider the function
$$f(x):=\log g(e^x)$$ and observe that it is additive, so that
$$f(x)=\log g(e^x)=ax,$$
$$g(e^x)=e^{ax},$$
$$g(x)=x^a.$$
Best Answer
A Geometric Proof
The posed inequality is equivalent to $\sin(x)\ge x-x^3/4.$
Consider the wedge of the unit circle below:
$\hspace{32mm}$
The area of the whole wedge (red and green regions) is $\frac12x$, and the area of the green triangle is $\frac12\sin(x)$. Thus, we get that $\sin(x)\le x$. Furthermore, the area of the red region is $\frac12(x-\sin(x))$.
Noting that the red region is contained in the rectangle with base $2\sin(x/2)$ and height $1-\cos(x/2)$, we get that $$ \begin{align} \tfrac12(x-\sin(x)) &\le2\sin(x/2)(1-\cos(x/2))\\ &=4\sin(x/2)\sin^2(x/4)\\ &\le x^3/8 \end{align} $$ which yields $$ x-x^3/4\le\sin(x) $$ as desired.