Is the following a proof of Euler's product formula for sine (see (8) below)? If no, then why?
The sine can be a polynomial by the taylor series for sine, which coverage for all $x \in \mathbb{R}.$
As known, the roots of this polynomial are $n\pi,~n \in \mathbb{Z}$.
Therefore, for some constants $C_n$, we can write this polynomial like that: $$\sin(x) = \prod_{n=-\infty}^{\infty}C_n\left(x-\pi n\right) = C_0x\prod_{n=1}^{\infty}C_{-n}C_n\left(x-\pi n\right)\left(x+\pi n\right) \tag{1}$$ Let $B_n = C_{-n}C_n$ and $B_0 = C_0$. Therefore, $$\frac{\sin\left(x\right)}{x}=B_0\prod_{n=1}^{\infty}B_n\left(x^2-\left(\pi n\right)^2\right) \tag{2}$$
As known, $\lim_{x\to 0}{\frac{\sin\left(x\right)}{x}} = 1$. Hence, $$\lim_{x\to 0}{\left [B_0\prod_{n=1}^{\infty}B_n\left(x^2-\left(\pi n\right)^2\right) \right ]} = B_0\prod_{n=1}^{\infty}\left ( -B_n\left(\pi n\right)^2 \right ) = 1 \tag{3}$$
Therefore, $$B_0=\frac{1}{\prod_{n=1}^{\infty}\left(-B_n\left(\pi n\right)^2\right)} \tag{4}$$
Finally, $$\begin{align}\sin\left(x\right)&=B_0x\prod_{n=1}^{\infty}B_n\left(x^2-\left(\pi n\right)^2\right) \tag{5}\\ &=\frac{x\prod_{n=1}^{\infty}B_n\left(x^2-\left(\pi n\right)^2\right)}{\prod_{n=1}^{\infty}\left(-B_n\left(\pi n\right)^2\right)} \tag{6}\\ &=x\prod_{n=1}^{\infty}\frac{B_n\left(x^2-\left(\pi n\right)^2\right)}{\left(-B_n\left(\pi n\right)^2\right)} \tag{7}\\ &=x\prod_{n=1}^{\infty}\left(1-\left(\frac{x}{\pi n}\right)^2\right) \tag{8}\end{align}$$
Can this be a proof? Thank you.
Definitely not.
Your argument relies only on the facts that the roots of $\sin x$ are $\pi n$, and that the $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$. There are plenty of functions satisfying these requirements. One may start with, say, $e^{x^2} \sin x $, and apply your reasoning equally well.
The reason it doesn't works I believe is the way you change the order of multiplications. To do so you need to prove first that those products converge absolutely. Otherwise it is as dangerous as changing the order of summation of a conditionally convergent series.