Given the curve $C = \{(x,y) \in \mathbb{R}^2: xy=1\}$, and given a point $(p,q)$ not on the curve, how do you find specific values for $\delta_1$, $\delta_2$ such that the open rectangle $R_{(p,q)}$ with vertices located at $(p \pm \delta_1, q \pm \delta_2)$ does not contain any points in $C$?
My Attempts at Finding $\delta_1$ and $\delta_2$:
At first I attempted to find a small square that would fit in $\mathbb{R}^2 - C$, by taking a point $(p,q)$ not on $C$, finding the intersections of the lines of slopes 1 and -1 through $(p,q)$, and I obtained the $x$-values for the intersection points. For example, for the line of slope 1 through $(p,q)$, you can obtain two intersection points $$x_{1,2} = \frac{(p-q) \pm \sqrt{(q-p)^2 + 4}}{2}.$$ I similarly would acquire $x_{3,4}$ such that the line of slope -1 through $(p,q)$ intersected $C$. For $i \in \{1,2,3,4\}$ and with $d_i = \|(x_i, \frac{1}{x_i}) - (p,q)\|$, I would then I would take $\delta_1 = \delta_2 = \frac{1}{4\sqrt{2}}\text{min}_{i \in \{1,2,3,4\}}d_i$. A special case would be made for points with $p = 0$ or $q = 0$.
Conceptually, this approach should give us a square with side length equal to $\delta$ with center $(p,q)$. However, I am unsatisfied, as I want to explicitly show that for any point $(u,v)$ in $R_{(p,q)}$, that $(u,v) \notin C$. I have also tried $\delta_1 = \delta_2 = \frac{1}{2\sqrt{2}}\text{min}(|q - \frac{1}{p}|,|p - \frac{1}{q}|)$. However I cannot seem to use triangle inequality gymnastics in any way to show that $(u,v)$ is not on the curve. There has got to be a simpler way to find explicit values for $\delta_1$ and $\delta_2$ in terms of the center point $(p,q)$ such that this rectangle lies in the complement of $C$, but I can't find it.
The problem amounts to an $\epsilon-\delta$ proof of the continuity of $(x,y)\mapsto xy$ at the point $(p,q)\notin C,$ for the max-norm on $\Bbb R^2.$ Indeed:
Let $$\epsilon:=|pq-1|>0.$$ By the triangular inequality, in order to produce a square centered at $(p,q)$ and disjoint from $C,$ it is sufficient to find some $\delta>0$ such that $$\left(|x-p|<\delta\text{ and }|y-q|<\delta\right)\implies|xy-pq|<\epsilon.$$
Now, if $|x-p|<\delta$ and $|y-q|<\delta,$ then $$|xy-pq|=|x(y-q)+q(x-p)|\le\left(|x|+|q|\right)\delta<\left(\delta+|p|+|q|\right)\delta$$ hence it suffices to take any $\delta>0$ such that $$\delta^2+\left(|p|+|q|\right)\delta-\epsilon\le0,$$ i.e. such that $$\delta\le\frac{-|p|-|q|+\sqrt{\left(|p|+|q|\right)^2+4\epsilon}}2.$$