We want to find the closest that the hyperbola $x^2-4y^2=4$ comes to one of the lines $y = \pm x$. This will occur at points on the hyperbola where the gradient is $\pm 1$ (as in the green line in the picture above). Differentiate the hyperbola equation to get $2x - 8yy' = 0$, and put $y'=\pm1$ to find that $x=\pm4y$. Substitute that into the hyperbola equation, getting $12y^2=4$, or $y = \pm1/\sqrt3$, $x = \pm4/\sqrt3$. Thus $|x| - |y| = 3/\sqrt3 = \sqrt3$.