Prove that the four points formed by the intersection of lines of the form $ax\pm by\pm c=0$ form a rhombus whose area is $\frac{2c^2}{|ab|}$
I found four lines as follows:
$ax+by+c=0\\ ax+by-c=0\\ ax-by+c=0\\ ax-by-c=0\\$
Now, what should I do next??
Your first and second lines are clearly parallel, as are the third and fourth lines. So they do not define intersection points.
There are four other pairs of lines, each pair defining an intersection point. They are 1st-3rd, 1st-4th, 2nd-3rd, 2nd-4th. Solve each pair of equations by any method (substitution, elimination, Cramer's rule, etc.). That gives you the coordinates of the four points.
Your next step is to find the length of the sides, which will turn out to be equal. This shows that the figure is a rhombus. Or, you could show that the diagonals are perpendicular. Since we already know the figure is a parallelogram, this also shows it to be a rhombus.
Last, find the area of the figure. Do you know the shoelace formula for finding the area of a polygon defined by the coordinates of its vertices?
Note that, for the problem to make sense, that it must be true that $a\ne 0$, $b\ne 0$, and $c\ne 0$. Otherwise you do note get four points or a rhombus at all.