I am trying to prove part $(i)$ here.
The following is my attempt at the problem is it correct?
Proof. Let $\alpha = (a,b)\in D$ and $\beta = (x,y)\in R_{(a,b)}$ and $O = (0,0)$, then given the description of $R_{(a,b)}$ in the above proposition we have $$R_{(a,b)} = \left\{(x,y)\in\mathbf{R}^2:|x-a|<\frac{1-r}{8\sqrt{2}}\text{ and }|y-b|<\frac{1-r}{8\sqrt{2}}\right\}$$ implying $$|x-a|^2 = (x-a)^2<\left(\frac{1-r}{8\sqrt{2}}\right)^2$$ $$|y-b|^2 = (y-b)^2<\left(\frac{1-r}{8\sqrt{2}}\right)^2$$ thus $$(x-a)^2+(y-b)^2<2\left(\frac{1-r}{8\sqrt{2}}\right)^2 = \left(\frac{1-r}{8}\right)^2<(1-r)^2$$ but then $$d(\alpha,\beta) = \sqrt{(x-a)^2+(y-b)^2}<\sqrt{(1-r)^2} = 1-r$$
appealing to the triangle inequality implies that
$$\sqrt{x^2+y^2} = d(O,\beta)\leq d(O,\alpha)+d(\alpha,\beta)<r+(1-r) = 1$$ the result in question then is evident.
$\blacksquare$
Given this i think it should be easy to prove that $D = \bigcup_{(a,b)\in D}R_{(a,b)}$.
Yes, it is correct. And, since you always have $R_{(a,b)}\subset D$, then$$D=\bigcup_{(a,b)\in D}\bigl\{(a,b)\bigr\}\subset\bigcup_{(a,b)\in D}R_{(a,b)}\subset D$$and therefore $\bigcup_{(a,b)\in D}R_{(a,b)}=D$ indeed.