(2) B Hint: The intersection points of the three lines are $A(2,-2)$, $B(2,-8)$, and $C(5, -5)$. Region $D$ is $\triangle ABC$. Point $A$ is the closest to the line $y=2x$, with a distance of $\frac{|2 \times 2-1 \times(-2)|}{\sqrt{5}}=\frac{6 \sqrt{5}}{5}$. Therefore, the closest distance between any two points in regions $D$ and $E$ is $\frac{12 \sqrt{5}}{5}$. Hence, the answer is B.
(2) The points $(x, y)$ in the plane that satisfy the constraints $\left\{\begin{array}{l}x \geqslant 2, \\ x+y \leqslant 0, \\ x-y-10 \leqslant 0\end{array}\right.$ form a region $D$. The region $D$ is symmetric to region $E$ with respect to the line $y=2 x$. Then, the distance between the closest points in regions $D$ and $E$ is ( ).
(A) $\frac{6 \sqrt{5}}{5}$
(B) $\frac{12 \sqrt{5}}{5}$
(C) $\frac{8 \sqrt{3}}{5}$
(D) $\frac{16 \sqrt{3}}{5}$