The distance from one corner to the opposite corner of a $d$-dimensional cube is always longer than the distance from one corner to the opposite corner of a $d-1$-dimensional cube. Suppose you have a piecewise-linear path from opposite corners of a $d$-diensional cube, i.e. from $(0, 0, ..., 0)$ to $(1, 1, ..., 1)$. Suppose your path is piecewise linear, suppose it consists of the straight-line path between some sequence of points $(0,0,...,0),(x_{1,1}, x_{1,2},...x_{1,d}), (x_{2,1}, x_{2,2},...x_{2,d}),$ $...,(x_{n,1},x_{n,2},...,x_{n,d}), (1,1,...1)$ for some $n$. Now consider the straight-line path between the alternative sequence of points $(0,0,...,0),(x_{1,1}, x_{1,2},...x_{1,d-1},0),$ $(x_{2,1}, x_{2,2},...x_{2,d-1},0),...,$ $(x_{n,1},x_{n,2},...,x_{n,d-1},0),$ $(1,1,...,1,0)$. This alternative path is obviously shorter. Thus, it is a shorter path between opposite sides of a $d-1$-dimensional cube.

EDIT: As has been pointed out in the comments, this isn't an answer to the original question. It only works in a single specific case.

The distance from one corner to the opposite corner of a $d$-dimensional cube is always longer than the distance from one corner to the opposite corner of a $d-1$-dimensional cube. Suppose you have a piecewise-linear path from opposite corners of a $d$-diensional cube, i.e. from $(0, 0, ..., 0)$ to $(1, 1, ..., 1)$. Suppose your path is piecewise linear, suppose it consists of the straight-line path between some sequence of points $(0,0,...,0),(x_{1,1}, x_{1,2},...x_{1,d}), (x_{2,1}, x_{2,2},...x_{2,d}),$ $...,(x_{n,1},x_{n,2},...,x_{n,d}), (1,1,...1)$ for some $n$. Now consider the straight-line path between the alternative sequence of points $(0,0,...,0),(x_{1,1}, x_{1,2},...x_{1,d-1},0),$ $(x_{2,1}, x_{2,2},...x_{2,d-1},0),...,$ $(x_{n,1},x_{n,2},...,x_{n,d-1},0),$ $(1,1,...,1,0)$. This alternative path is obviously shorter. Thus, it is a shorter path between opposite sides of a $d-1$-dimensional cube.