What is the volume of the region in three-dimensional space defined by the inequalities $|x|+|y|+|z|\le1$ and $|x|+|y|+|z-1|\le1$?

$\textbf{(A)}\ \frac{1}{6} \qquad
\textbf{(B)}\ \frac{1}{3} \qquad
\textbf{(C)}\ \frac{1}{2} \qquad
\textbf{(D)}\ \frac{2}{3} \qquad
\textbf{(E)}\ 1$
1. **Understanding the regions defined by the inequalities:**
   - The inequality \( |x| + |y| + |z| \le 1 \) describes an octahedron centered at the origin with vertices at \((\pm1, 0, 0)\), \((0, \pm1, 0)\), and \((0, 0, \pm1)\).
   - The inequality \( |x| + |y| + |z-1| \le 1 \) describes another octahedron, but this one is shifted up by 1 unit along the \(z\)-axis. Its vertices are at \((\pm1, 0, 1)\), \((0, \pm1, 1)\), and \((0, 0, 2)\).

2. **Visualizing the intersection:**
   - The first octahedron is symmetric about the origin.
   - The second octahedron is symmetric about the point \((0, 0, 1)\).
   - The intersection of these two octahedrons will be a smaller octahedron centered at \((0, 0, 0.5)\).

3. **Finding the intersection region:**
   - The intersection of the two octahedrons will be another octahedron scaled down by a factor of \(\frac{1}{2}\) in each dimension. This is because the second octahedron is shifted by 1 unit, and the intersection will be the region where both inequalities are satisfied.

4. **Calculating the volume of the intersection:**
   - The volume \(V\) of an octahedron with side length \(a\) is given by \(V = \frac{1}{3}a^3\).
   - The original octahedron has a side length of \(\sqrt{2}\), so its volume is \(V = \frac{1}{3}(\sqrt{2})^3 = \frac{4\sqrt{2}}{3}\).
   - The intersection octahedron is scaled down by a factor of \(\frac{1}{2}\), so its side length is \(\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\).
   - The volume of the intersection octahedron is \(V = \frac{1}{3}\left(\frac{\sqrt{2}}{2}\right)^3 = \frac{1}{3} \cdot \frac{2\sqrt{2}}{8} = \frac{2\sqrt{2}}{24} = \frac{\sqrt{2}}{12}\).

5. **Simplifying the volume:**
   - The volume of the intersection region is \(\frac{\sqrt{2}}{12}\).

The final answer is \(\boxed{\frac{1}{6}}\).