Prove that $\{(x,y)\mid xy>0\}$ is open I need to prove this using open balls. So the general idea is to construct a open ball around a point of the set. A point $(x,y)$ such that $xy>0$. Then we must prove that this ball is inside the set. However, I don't know how to find a radius for this open ball. Can somebody help me in this proof?

The region $xy>0$ is just the union of (1) the interior of the first quadrant, and (2) the interior of the third quadrant.  That is the set you're trying to prove to be open.
If $(x,y)$ is in the first quadrant, it distance from the $x$-axis is $y$ and its distance from the $y$-axis is $x$.  Make the radius of the disk less than or equal to both of those and then the disk won't intersect either of the boundaries, but rather will remain within the open set.  The same thing works in the third quadrant except that $x$ and $y$ are negative so you need $|x|$ and $|y|$.  So $|x|$ and $|y|$ do it, for both quadrants.
