The area of the closed region bounded by the equation | x | + | y | = 2 in the two-dimensional plane is

The correct option is C.

The given equation is $\left|x\right|+\left|y\right|=2$.

This equation represents a closed region in the 2D Cartesian plane. We can determine the boundary lines by considering the signs of $x$ and $y$ in each of the four quadrants:

1. In the first quadrant ($x\ge 0,y\ge 0$): $x+y=2$
2. In the second quadrant ($x\le 0,y\ge 0$): $-x+y=2$
3. In the third quadrant ($x\le 0,y\le 0$): $-x-y=2$ (or $x+y=-2$)
4. In the fourth quadrant ($x\ge 0,y\le 0$): $x-y=2$

Plotting these lines, we find they intersect the axes at:
- x-axis: (2, 0) and (-2, 0)
- y-axis: (0, 2) and (0, -2)

The region bounded by these lines is a square (or a rhombus) whose diagonals lie along the coordinate axes.

The length of each diagonal is:
$d_1=4$ (along the x-axis from -2 to 2)
$d_2=4$ (along the y-axis from -2 to 2)

The area of a rhombus/square given its diagonals is:
$\text{Area}=\frac{1}{2}\times d_1\times d_2=\frac{1}{2}\times 4\times 4=8$