From a square with sides of length $2m$, corners are cut away so as to form a regular octagon. What is the area of the octagon in $m^2$?
$\textbf{(A)}~2\sqrt3$
$\textbf{(B)}~\frac4{\sqrt3}$
$\textbf{(C)}~4\left(\sqrt2-1\right)$
$\textbf{(D)}~\text{None of the above}$
1. Let the side length of the square be \(2m\). We need to cut off the corners to form a regular octagon. Each corner cut will be an isosceles right triangle with legs of length \(x\).

2. The length of the side of the octagon can be determined by subtracting twice the leg length of the triangles from the side length of the square:
   \[
   2m - 2x
   \]

3. The diagonal of the isosceles right triangle (which is the hypotenuse) is given by:
   \[
   \sqrt{2}x
   \]

4. The length of the side of the octagon is also equal to the length of the side of the square minus twice the leg length of the triangles:
   \[
   2m - 2x = \sqrt{2}x
   \]

5. Solving for \(x\):
   \[
   2m - 2x = \sqrt{2}x
   \]
   \[
   2m = x(\sqrt{2} + 2)
   \]
   \[
   x = \frac{2m}{\sqrt{2} + 2}
   \]

6. Rationalizing the denominator:
   \[
   x = \frac{2m}{\sqrt{2} + 2} \cdot \frac{\sqrt{2} - 2}{\sqrt{2} - 2} = \frac{2m(\sqrt{2} - 2)}{2 - 4} = \frac{2m(\sqrt{2} - 2)}{-2} = m(2 - \sqrt{2})
   \]

7. The area of the original square is:
   \[
   (2m)^2 = 4m^2
   \]

8. The area of one of the triangles cut off is:
   \[
   \frac{1}{2}x^2 = \frac{1}{2}(m(2 - \sqrt{2}))^2 = \frac{1}{2}m^2(4 - 4\sqrt{2} + 2) = \frac{1}{2}m^2(6 - 4\sqrt{2}) = 3m^2 - 2\sqrt{2}m^2
   \]

9. Since there are 4 such triangles, the total area cut off is:
   \[
   4(3m^2 - 2\sqrt{2}m^2) = 12m^2 - 8\sqrt{2}m^2
   \]

10. The area of the octagon is the area of the square minus the area of the triangles:
    \[
    4m^2 - (12m^2 - 8\sqrt{2}m^2) = 4m^2 - 12m^2 + 8\sqrt{2}m^2 = -8m^2 + 8\sqrt{2}m^2 = 8m^2(\sqrt{2} - 1)
    \]

Thus, the area of the octagon is:
\[
\boxed{4(\sqrt{2} - 1)}
\]