Elementary proof that the MacLaurin series of $\sin x$ converges to $\sin x$ for all $x$ In my book it is given: 
$\sin x = x- \dfrac {x^3}{3!}+\dfrac{x^5}{5!}- \dfrac{x^7}{7!}...$
I googled around for a proof but couldn't understand any of them. I would like to know if there's any elementary high school level proof the series

A proper proof would require ideas from analysis/calculus and you should get familiar with these (if not already). The typical proof is via Taylor's theorem. A non-rigorous version uses the formula $$\sin nx=n\sin x-\frac{n(n^2-1^2)}{3!}\sin^3x+\frac{n(n^2-1^2)(n^2-3^2)}{5!}\sin^5x-\cdots$$ (convince yourself that the above formula is true by trying out odd values of $n$ like $3,5$) and then puts $nx=t$ where $t$ is constant, $n\to\infty, x\to 0$ to get $$\sin t=t-\frac{t^3}{3!}+\dots$$
Newton used another procedure to obtain infinite series for $\sin x$. Using the geometric definition of $\sin x$ (as mentioned in your comment) one obtains $$x=\int_{0}^{\sin x}\frac{dt} {\sqrt{1-t^2}}$$ for $x\in[-\pi/2,\pi/2]$ and expanding the integrand via binomial theorem and integrating term by term we get $$x=\sin x+\frac{1}{2}\frac{\sin^3x}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{\sin^5x}{5}+\dots$$ and then assuming $$\sin x=ax+bx^3+cx^5+\dots$$ and comparing coefficients we can easily find $a, b, c, d$. 
