Area of the region bounded by |x| + |y|≤ 2 is:

The correct option is 8.

To find the area of the region bounded by the inequality $\left|x\right|+\left|y\right|\le 2$, we can analyze the boundary of the region given by the equation:
$\left|x\right|+\left|y\right|=2$

Depending on the signs of $x$ and $y$ in the four quadrants, this equation represents four different straight lines:

1. In the first quadrant ($x\ge 0,y\ge 0$):
$x+y=2$

2. In the second quadrant ($x<0,y\ge 0$):
$-x+y=2$

3. In the third quadrant ($x<0,y<0$):
$-x-y=2$

4. In the fourth quadrant ($x\ge 0,y<0$):
$x-y=2$

Plotting these lines on the coordinate plane, we get a square whose vertices are:
$(2,0)$, $(0,2)$, $(-2,0)$, and $(0,-2)$.

The region $\left|x\right|+\left|y\right|\le 2$ represents the interior and boundary of this square.

The length of each diagonal of this square is the distance between opposite vertices, which is:
$d=2-(-2)=4$

The area of a square with diagonals of length $d$ is given by the formula:
$\text{Area}=\frac{1}{2}\times d^2$

Substituting $d=4$ into the formula:
$\text{Area}=\frac{1}{2}\times 4^2=\frac{1}{2}\times 16=8$

Alternatively, the square is composed of four identical right-angled triangles (one in each quadrant), each having a base of 2 units and a height of 2 units.
The area of one such triangle is:
$\text{Area of one triangle}=\frac{1}{2}\times 2\times 2=2$

Therefore, the total area of the region is:
$\text{Total Area}=4\times 2=8$