First, from
$$\begin{cases} x+2y>0, \\ x-2y>0, \\ (x+2y)(x-2y)=4, \end{cases}$$
we obtain
$$\begin{cases} x > 2|y| \ge 0, \\ x^2 - 4y^2 = 4. \end{cases}$$

By the symmetry, there is no loss of generality in considering only the case when $y \ge 0$. In view of $x > 0$, we need to find the minimum value of $x-y$ only.
Setting $u = x-y$, and substituting it into $x^2-4y^2=4$, we obtain
$$3y^2 - 2uy + (4 - u^2) = 0. \quad (*)$$
Equation (*) with respect to $y$ has real solutions. So we have
$$\Delta = 4u^2 - 12(4 - u^2) \ge 0.$$
Thereby
$$u \ge \sqrt{3}.$$

In addition, when $x = \frac{4}{3}\sqrt{3}$ and $y = \frac{\sqrt{3}}{3}$, we have $u = \sqrt{3}$.
Therefore, the minimum value of $|x| - |y|$ is $\sqrt{3}$.