$sin(x)$ is continuous, so it is uniform on closed and limited intervals. Let this interval be $I = [0,3\pi]$, so there exists a $\delta_0$ for a given $\epsilon$.
Then, let us consider all $x,y$ in $\mathbb{R}$. We can assume $y \le x$, and I claim $\delta = \pi$ will work (assuming that the $\delta_0$ does not work).
Since we want to take use of sine's period, I want to express $x,y$ in terms of $2k\pi$. So we can say that $2k\pi \le y \le 2k\pi + 2\pi \Rightarrow 2k\pi \le y \le x \le 2k\pi + 3\pi$. Rewrite to $$0 \le x- 2k\pi \le y-2k\pi \le 3\pi$$ Then set $a = x- 2k\pi$ and $b = y - 2k\pi$, and these are members of $I$ so sine at $x,y$ is equivalent to sin at $a,b$ who are less than $\epsilon$ apart if $\delta$ is 1 or $\delta_0$ (the smaller one is chosen).
does above work, or have I made some mistakes?
Hint:
Much simpler, in my opinion:
$$\left|\frac{\sin x-\sin y}{x-y}\right|=|(\sin c)'|=|\cos x|\le 1\implies |\sin x-\sin y|\le |x-y|$$
What theorem was used? Observe that the above works for any $\;x,y\in\Bbb R\;$