The distance to beat is the main diagonal from (0,0,0) to (1,1,1) which is sqrt(3) and right from the start I suspect this is the answer.

Want to know how far is the perpendicular distance from that diagonal line to any of the other vertices; then double it.

(Could do an axis modification so the opposite vertices are at (0,0,0) and (0,0,sqrt(3)) and look at the x,y coordinates of one of the other vertices.  I don't recall how to do a 3D axis rotation, so I'll try the other way.)

3D line of the main diagonal:  x=y=z

A vertex is (0,0,1), closest point on diagonal line is (a,a,a)

minimize dist = √((a-0)^2 + (a-0)^2 + (a-1)^2)

dist^2 = 3a^2 - 2a + 1

min if 6a - 2 = 0, a = 1/3.

dist = √((1/3)^2 + (1/3)^2 + (2/3)^2)

dist = √(6/9) = √6/3

2*dist = 2√6/3

So if we now imagine the cube rotated so the main diagonal is vertical, the widest horizontal distance is:

2√6/3 = 1.6329931618554518  which is less than √3.  In this rotated configuration, where opposite vertices are at (0,0,0) to (√3,√3,√3) there are 2 wide cross sections, at z heights of √3/3 and 2√3/3.

If we take a frontal plane through this figure, call the axes u and z, there will be extreme points at  (u,z) coordinates of:

(-√6/3, √3/3) and (√6/3, 2√3/3).

That distance is √( (2√6/3)^2  +  (√3/3)^2 )

  = √( (24/9)  +  (3/9) )

  = √(27/9) = √3  which should not surprise me because those two edges of the solid are exactly the result of 2 opposite vertices rotating.

So:  √3