Section of a 4-dimensional cube Consider 4-dimensional cube: $-1 \le x_1 \le 1, -1 \le x_2 \le 1,-1 \le x_3 \le 1, -1 \le x_4 \le 1$.
Could you show an octagon that is a section of this cube?
 A: What follows is not a section by an hyperplane, but a skew polygon, whose vertices are the columns $C_{k}$ of this matrix: 
$$M=\begin{pmatrix}1&0&-1&0&-1&0&1&0\\0&1&0&-1&0&-1&0&-1\\1&0&1&0&-1&0&-1&0\\0&1&0&1&0&-1&0&-1\end{pmatrix}$$
Proof: $\forall k=1 \cdots 7, \text{dist}(C_k,C_{k+1})=\sqrt{1^2+1^2+1^2+1^2}=2$ and dist($C_8,C_{1})=2$ as well.
It is a skew polygon.
Moreover, if one wants other such octagons, one can premultiply $M$ by a well chosen orthogonal matrix, for example, a permutation matrix.
A: Yes, it is possible.
First observe that it is possible to cut a cube in a way the cut is a regular hexagon.
Now extend that cube to the fourth dimension to get a tesseract (4D cube), and at the same time extend the cutting plane in the same direction to give a cutting hyperplane. This of course just adds an orthogonal dimension to the cut, so we get as cut figure a hexagonal prism.
Now we can cut the hexagonal prism with a plane so that of both hexagons, a part of a corner is cut away, so the cut is an octagon. That octagon is, of course, not regular, but that was not required.
This octagon is then, of course, also the cut of the two-dimensional plane with the four-dimensional tesseract.
For the coordinates of that polygon:
First, a hexagonal cut through a cube is, for example, given by the corners
$$(0,-1,-1), (-1,0,-1), (-1,1,0), (0,1,1), (1,0,1), (1,-1,0)$$
It is easily checked that those are, indeed, in a plane.
Extruding to the fourth dimension a hexagonal prisma with the hexagonal sides
$$(0,-1,-1,-1), (-1,0,-1,-1), (-1,1,0,-1), (0,1,1,-1), (1,0,1,-1), (1,-1,0,-1)$$
and
$$(0,-1,-1,1), (-1,0,-1,1), (-1,1,0,1), (0,1,1,1), (1,0,1,1), (1,-1,0,1)$$
Again, it is readily checked that those points all lie in a three-dimensional space. Also, it is just as easily checked that they indeed lie on the border of the tesseract.
Now all that remains is to cut through this with a plane. Let's take the plane that cuts off the corners at $(0,-1,-1,-1)$ and $(0,1,1,1)$ by cutting the adjacent sides of the respective hexagons in the middle. This condition fixes the  four points
$$(\frac{1}{2},-1,-\frac{1}{2},-1),
(-\frac{1}{2},-\frac{1}{2},-1,-1),
(-\frac{1}{2},1,\frac{1}{2},1),
(\frac{1}{2},\frac{1}{2},1,1)$$
The other four points of the octagon are then found by intersecting the plane through those points with the four lines connecting related non-cut off corners of the two hexagons.
If I didn't miscalculate, they should be
$$(-1,0,-1,-\frac13), (-1,1,0,\frac13), (1,0,1,\frac13), (1,-1,0,-\frac13)$$
