Consider a face $F$ of the cube, and cut the cube with a plane $P$ that is perpendicular to $F$.
For example, if $F$ is a horizontal face, then $P$ is a vertical plane. We explore the cross section of the cube that is cut by $P$.
Consider $xyz$ coordinates in space.
We may take without loss of generality the unit cube with one vertex at the origin and the opposite vertex that is $(1,1,1)$. Moreover, we may suppose that $F$ is the horizontal face consisting of the points whose last coordinate is $0$. In the $xy$-plane, the points of $F$ become the points of the unit square $S$, while the vertical plane $P$ cuts the $xy$-plane in a line $L$.
If the intersection $S\cap L$ is known, what is the intersection of $P$ and the cube?
The intersection of $P$ and the cube is the Cartesian product $(S\cap L)\times [0,1]$. We prove the equality of these two sets by showing that any of them is contained in the other.
We first show that the points in $(S\cap L)\times [0,1]$ are contained both in the cube and in the plane $P$: they are contained in the cube because, thanks to $S$, the three coordinates are between $0$ and $1$; they are contained in $P$ because (by definition of $L$) this is the case for the points in $L\times \{0\}$, and a vertical plane contains the vertical lines through its points.
To show the other inclusion, we prove that a point $X=(a,b,c)$ in the intersection of $P$ and the cube is such that $(a,b)$ is a point in $S\cap L$. Since $X$ is a point of the cube, both $a$ and $b$ are between $0$ and $1$ and hence $(a,b)$ is in $S$. Since $X$ is in the vertical plane $P$, also the point $(a,b,0)$ is in $P$, which gives that $(a,b)$ is in $L$.
What are the geometric figures that can be the intersection between the square $S$ and the line $L$?
The intersection $S\cap L$ could be empty (for example, if $L$ is the line given by $x=2$), and now we can exclude this case.
Since $S$ is bounded, the intersection $S\cap L$ will be a bounded set of the plane. Since $S$ and $L$ are convex sets, the same holds for $S\cap L$. Thus the set $S\cap L$, being a bounded, non-empty and convex subset of the line $L$, can only be a point or a segment.
The intersection $S\cap L$ can consist in one point (one of the vertices of $S$) and it can consist in a segment (for example, one side of $S$).
Prove that, if $S\cap L$ is a segment, then its endpoints are on the sides of $S$. Moreover, if the two endpoints belong to the same side of $S$, then $S\cap L$ is precisely this side.
If the endpoints belong to the same side of $S$, then this side is fully contained in the line $L$ and we conclude (in this case, the endpoints are square vertices).
Now suppose that $S\cap L$ contains an interior point $X=(x_0,y_0)$ of the square. We conclude by proving that $X$ is not an endpoint for $S\cap L$. There exist two real numbers $r,r'$ such that the points $X_t=(x_0+rt,y_0+r't)$ belong to $L$ for every real number $t$. If $t$ is sufficiently close to $0$, then the coordinates of $X_t$, like the ones of $X$, will be strictly between $0$ and $1$ and hence $X_t$ is in $S\cap L$, showing that $X$ cannot be an endpoint for this segment.
Prove that, if $S\cap L$ is a point, then it is a vertex of the square $S$.
The intersection point cannot be an interior point of the square because, as shown before, there would be neighboring points that belong both to the cube and the line.
Now suppose that the intersection point is on a square side but it is not a vertex. Without loss of generality let this point be of the form $(x_0,0)$ for some real number $x_0$ that is between $0$ and $1$. If $L$ is horizontal, then $S\cap L$ would also contain $(0,0)$, while if it is vertical, it would also contain $(x_0,1)$, contradicting that $S\cap L$ consists of a single point. Now suppose that $L$ is neither vertical nor horizontal, so there is a non-zero real number $m$ such that the points $(x_0+t,mt)$ are contained in $L$ for all real numbers $t$. Taking $t$ sufficiently close to $0$ and such that $mt$ is strictly positive, the point $(x_0+t,mt)$ is a second point in $S\cap L$, contradiction.
What are the possible intersections between the cube and the plane $P$?
Suppose that the intersection is non-empty. Then $S\cap L$ is non-empty and hence it is either a square vertex or a segment whose endpoints belong to sides of $S$. In the former case, the intersection between $P$ and the cube is one of the vertical edges of the cube. In the latter case, the intersection is a rectangle.
The vertical side of the rectangle has length $1$, while the horizontal side of the rectangle has a length that is strictly positive and at most $\sqrt{2}$. (Indeed, the distance between two distinct points on the sides of $S$ can be any positive number that is at most the length of the square diagonal.)
The area of the rectangle can then be any positive real number that is at most $\sqrt{2}$.
Remark: If we had a cube with side $\ell$ we should rescale the side lengths by $\ell$ and the area by $\ell^2$.
When do we obtain a square as the intersection between the cube and the plane $P$?
The requested condition means that the intersection $S\cap L$ is a segment of length one. This is possible if $L$ is parallel to a side of $S$ (and $S\cap L$ is non-empty), which means that $P$ is parallel to a cube face and intersects the cube.
Alternatively, the two endpoints of the segment $S\cap L$ must be on neighboring sides of $S$ and at distance $1$. If we fix a point $A$ on a side of $S$, and a neighboring side, then there is precisely one point $B$ on this side that is at distance $1$ from $A$.
