Brazil

The answer is $\frac{3\sqrt{2}}{4}$. Just consider the points $(\frac{3}{4},0,0)$, $(0,\frac{3}{4},0)$, $(1,\frac{1}{4},1)$ and $(\frac{1}{4},1,1)$.

Now suppose that there is a square with side $\ell >\frac{3\sqrt{2}}{4}$ inside the cube. First, it's not hard to prove that we can suppose without loss of generality that the centers of the cube and of the square coincide. If it isn't the case, consider the three planes that cut the cube in two blocks with dimensions $\frac{1}{2},1,1$. Now focus on only one of the planes, which we'll call $\alpha$. If the center of the square is not in $\alpha$, it is inside one of the two blocks defined by the plane. Next, draw a plane $\beta$ passing through the center of the square and parallel to $\alpha$. It cuts the square in two congruent pieces and the cube in two blocks, one smaller than the other. Half of the square is inside the smaller block, so if we perform a translation that transforms $\beta$ in $\alpha$ there wouldn't be any problems, that is, the square won't go outside the cube. Repeating this procedure two more times with the two other planes, the center of the square will eventually coincide with the center of the cube.

The next step is to draw a sphere $S$ with center on the center of the square (which now coincides with the center of the cube) and radius $\frac{\ell \sqrt{2}}{2}>\frac{3}{4}$. Note that all vertices of the square will lie on the surface of $S$ and that two opposite vertices of the square will be antipodal on $S$. Then we know where are the vertices of the square: each is in one of the $8$ regions determined by the intersection of the surface of $S$ and the cube. One vertex is in one of these regions, say $R$; the other one lies on the opposite region, say $T$.

Now consider a block containing two neighbour regions. This block has dimensions $1,x$ and $x$. Recalling that $S$ has radius bigger than $\frac{3}{4}$, then $x$ is less than $\frac{1}{2}-\sqrt{(\frac{3}{4}{)}^2-(\frac{\ell \sqrt{2}}{2}{)}^2}=\frac{1}{4}$. But then the maximum distance between any two points inside the block is less than $\sqrt{1^2+2x^2}=\frac{3\sqrt{2}}{4}$, so it is impossible to have two vertices of the square in two neighbour regions. But $R$ eliminates three regions and $T$ eliminate the other three remaining regions. The other two vertices of the square won't have any region to go, contradiction.

Thus the biggest square contained in the cube has side $\frac{3\sqrt{2}}{4}$.