Starting off where you finished and credits to this pdf:
$$\sin'(x) = \lim_{\Delta x \to 0} \left(\cos x \frac{\sin \Delta x}{\Delta x} + \sin x \frac{\cos \Delta x - 1}{\Delta x} \right)$$ $$= \cos x \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} - \sin x \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x}$$
Part I: prove that $\displaystyle \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1$.
Assuming that $OA = 1$ (i.e. it's a unit circle), we have $\sin \theta = BC$. We then note that
[Area of triangle $AOD$] > [Area of sector $AOC$] > [Area of triangle $AOC$]
which means
$$\frac{1}{2} \frac{\sin \theta}{\cos \theta} > \frac{1}{2} \theta > \frac{1}{2} \sin \theta$$
From this we get $$\cos \theta < \frac{\sin \theta}{\theta} < 1$$
and since $\displaystyle \lim_{\theta \to 0} \cos \theta = 1$, we have that $\displaystyle \lim_{\theta \to 0} = 1$.
Part II: prove that $\displaystyle \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = 0$.
$$\lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = \lim_{\Delta x \to 0} \frac{(1 - \cos \Delta x) (1 + \cos \Delta x)}{(1 + \cos \Delta x)\Delta x} = \lim_{\Delta x \to 0} \frac{1 - \cos^2 \Delta x}{(1 + \cos \Delta x)\Delta x}$$
$$ = \lim_{\Delta x \to 0} \frac{\sin^2 \Delta x}{(1 + \cos \Delta x)\Delta x} = \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} \frac{\sin \Delta x}{1 + \cos \Delta x}$$
We have proven in part I that $\displaystyle \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1$, and we note that $\displaystyle \frac{\sin \Delta x}{1 + \cos \Delta x} = \frac{0}{1+1} = 0$.
We finally put all this together to get that $$\boxed{\displaystyle \sin'(x) = \cos x \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} - \sin x \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = \cos x \cdot 1 - \sin x \cdot 0 = \cos x}$$
[This answer has significant overlap with my answer to a related but different question, Intuitive understanding of the derivatives of $\sin x$ and $\cos x$. ]
Disclaimer: I am not sure how to make the following totally rigorous, but it uses the unit circle definition and avoids using the limit in question, while seeming pretty convincing, hopefully enough to be worthy of an answer.
The curve $\gamma(t)=(\cos(t),\sin(t))$ has everywhere unit length and unit speed, as follows readily from the definition. $\|\gamma(t)\|^2=\cos^2(t)+\sin^2(t)=1$ everywhere by definition, because $(\cos(t),\sin(t))$ is on the unit circle. $\|\gamma'(t)\|=1$ everywhere because it is rate of change of the distance traveled around the circle with respect to $t$, and $t$ is the distance traveled by definition of radian (and we are using radians here, or else the result would be false).
From $\|\gamma(t)\|^2=\gamma(t)\cdot \gamma(t)\equiv 1$ we conclude from the product rule that $\gamma'(t)\cdot \gamma(t)\equiv 0$, so that $\gamma'(t)$ is a unit vector perpendicular to $(\cos(t),\sin(t))$. There are only two such vectors in the plane, namely $(-\sin(t),\cos(t))$ and $(\sin(t),-\cos(t))$, but it is easy to rule out the latter, for instance by noticing where $\sin$ is increasing and decreasing. Hence $\gamma'(t)=(-\sin(t),\cos(t))$ for all $t$, and in particular the derivative of $\sin$ is $\cos$.
Remark: As far as I know the most significant gap in the above is the rigorous justification that $\gamma$ is even differentiable (which is equivalent to knowing that $\cos$ and $\sin$ are differentiable).
Added: It can be shown that $\gamma$ is differentiable without further assumptions about trig functions. There is probably a nicer way to do so than what follows, but the following is something. Note that we are dealing with the unit speed parametrization of the curve $x^2+y^2=1$. Near $(1,0)$ for example, we have $x=\sqrt{1-y^2}$, so we can parametrize the curve smoothly as $\alpha(y) = (\sqrt{1-y^2}, y)$. Note that $\alpha$ is smooth on $-1<y<1$ and is tracing the right half of the circle. We can then reparametrize with respect to arclength as outlined here. The arclength along the circle from $(1,0)$ to $\alpha(y)$ with $y>0$ is the radian measure $\theta$ of the angle (by definition of radian), so $\theta(y)=\int_0^y\|\alpha'(t)\|\,dt=\int_0^y\frac{1}{\sqrt{1-t^2}}\,dt$ (which is $\arcsin(y)$). Note that $\theta(y)$ is smooth, hence so is the inverse function $y(\theta)=\sin(\theta)$ by the inverse function theorem.
More generally, I think something along these lines, with the implicit and inverse function theorems, can be used to show that algebraic curves in the plane have smooth unit speed parametrizations away from singular points, but my attempts to search for a better statement and reference for such claims have been unfruitful.
Best Answer
We define the sine function, $\sin(x)$, as the inverse function of the function $f(x)$ given by
for $|x|< 1$.
It is straightforward to show that since $\frac{1}{\sqrt{1-t^2}}$ is positive and continuous for $t\in (-1,1)$, $f(x)$ is continuous and strictly increasing for $x\in (-1,1)$ with $\displaystyle\lim_{x\to 0}f(x)=f(0)=0$.
Therefore, since $f$ is continuous and strictly increasing, its inverse function, $\sin(x)$, exists and is also continuous and strictly increasing with $\displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0$.
From $(1)$, we have the bounds (SEE HERE)
for $x\in (-1,1)$, whence applying the squeeze theorem to $(2)$ yields
$$\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3$$
Finally, let $y=f(x)$ so that $x=\sin(y)$. As $x\to 0$, $y\to 0$ and we can write $(3)$ as
$$\lim_{y\to 0}\frac{y}{\sin(y)}=1$$
from which we have
as was to be shown!