Applying the reverse construction technique leads to \begin{align*} &\sin^2 20^\circ -\sin 5^\circ (\sin 5^\circ +\frac{\sqrt{6}-\sqrt{2}}{2}\cos 20^\circ)\\ =\quad & \sin^2 20^\circ -\sin 5^\circ (\sin 5^\circ +2\sin 15^\circ\cos 20^\circ)\\ =\quad & \sin^2 20^\circ - \sin 5^\circ(\sin 5^\circ + \sin 35^\circ - \sin 5^\circ)\\ =\quad & \sin^2 20^\circ - \sin 5^\circ \sin 35^\circ\\ =\quad & \frac{1-\cos 40^\circ}{2} - \frac{1}{2}\cdot(\cos 30^\circ - \cos 40^\circ)\\ =\quad & \frac{1}{2}\cdot\Big(1-\frac{\sqrt{3}}{2}\Big)\\ =\quad & \boxed{\frac{2-\sqrt{3}}{4}} \end{align*}