This question requires us to determine the area enclosed by the given curves:
The curves \( |y + x| \le 1 \) and \( |y - x| \le 1 \) form a square with vertices at points where the lines intersect. Let's find these vertices:
From \( |y + x| \le 1 \), we get two lines: \( y + x = 1 \) and \( y + x = -1 \).
From \( |y - x| \le 1 \), we also get two lines: \( y - x = 1 \) and \( y - x = -1 \).
By solving these pairs of equations, we find the intersection points (vertices of the square):
Solving \( y + x = 1 \) and \( y - x = 1 \), we find that \( x = 0 \) and \( y = 1 \), so one vertex is \( (0, 1) \).
Solving \( y + x = 1 \) and \( y - x = -1 \), we find that \( x = 1 \) and \( y = 0 \), so another vertex is \( (1, 0) \).
Solving \( y + x = -1 \) and \( y - x = 1 \), we find that \( x = -1 \) and \( y = 0 \), so another vertex is \( (-1, 0) \).
Solving \( y + x = -1 \) and \( y - x = -1 \), we find that \( x = 0 \) and \( y = -1 \), so the last vertex is \( (0, -1) \).
The square thus formed has side length 2, as each line is two units apart. Therefore, the area of this square is \( \text{side}^2 = 2^2 = 4 \) square units.
Considering the ellipse given by \( 2x^2 + 2y^2 = 1 \):
This can be rewritten as \( x^2 + y^2 = \frac{1}{2} \), an equation for a circle with radius \( \frac{1}{\sqrt{2}} \).
The area of the circle is \( \pi \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{\pi}{2} \) square units.
We want the area common to both the square and the circle:
The desired area is the area of the square minus the area of the circle inside it, because the circle is entirely contained within the square.
Thus, the enclosed area is \( 4 - \frac{\pi}{2} \) square units.
Hence, the correct answer is \( \left(2 - \frac{\pi}{2}\right) \) sq units.