6. In the Cartesian coordinate plane $x O y$, the area of the region determined by the system of inequalities
$$
\left\{\begin{array}{l}
|x| \leqslant 2, \\
|y| \leqslant 2, \\
|| x|-| y|| \leqslant 1
\end{array}\right.
$$

is $\qquad$
6. 12 .

Obviously, the region is symmetric with respect to both the $x$-axis and the $y$-axis. Therefore, we can assume $x \geqslant 0, y \geqslant 0$.

From $\left\{\begin{array}{l}0 \leqslant x \leqslant 2, \\ 0 \leqslant y \leqslant 2, \\ -1 \leqslant x-y \leqslant 1\end{array}\right.$, draw the part of the region in the first quadrant (including the coordinate axes), and then symmetrically draw the entire region, as shown in Figure 4.

The area of the region sought is the area of a square with side length 4 minus the area of eight isosceles right triangles with leg length 1, i.e., $4^{2}-8 \times \frac{1}{2}=12$.