The result follows easily from the following theorem, which we will prove.
Theorem wrote: The perimeter of a convex planar polygon equals [imath]\pi[/imath] times the average length of its orthogonal projections over all directions.
In other words, imagine putting the Sun infinitely far away in some direction, and finding the length of the line segment that is the polygon's shadow falling onto a line perpendicular to the sun's rays. The polygon's perimeter is proportional to the length of this shadow averaged over all directions that the Sun could in, which correspond to angles from 0 to [imath]2 \pi[/imath]. (I've forgotten the name of this result, it's called some famous guy's theorem; I'd be obliged to anyone who can find a reference. Also, I think this result, if extended to convex, bounded planar regions, implies that every such region has a well-defined and finite perimeter.)
This theorem implies the result, since if A lies inside B, then every shadow of A lies within the corresponding shadow of B, and so the average shadow size of A is at most that average shadow size of B, and therefore the perimeter of A is at most that of B.
Now we'll prove the theorem. I'll try to use words rather than symbols so as to be non-technical.
First, we note that if we orthogonally project a line segment in the plane along a fixed direction (i.e. take its shadow from an infinitely far light source onto a line perpendicular to the light rays), the shadow's length scales with the segment's length. So, the average length of a projection in a random direction is proportional to the segment's length. We could find the constant of proportionality by integrating over all angles, but instead will defer finding it until later using a trick.
Now, the perimeter of a polygon is made of line segments. By linearity of expectation, the average total length (over all projection directions) of the shadows of the polygon's side (taken separately) equals the sum of the each side's average shadow-length, which is proportional to the length of the side. Therefore, the average total shadow length of the sides is proportional to the sum of their lengths, which is the polygon's perimeter.
Next, we note that the shadow of the polygonal region is covered by the shadows of its sides. In fact, since each ray from the Sun passes through the polygon twice, the shadows of the sides doubly cover the shadow of the polygon. So, the length of the shadow of the polygon is half the length of the shadows of the sides. Averaging over the shadows gotten from all directions, we find that the average length of the shadow of the polygon is proportional to the polygon's perimeter.
We can find the constant of proportionality by a single example. We see that the results should hold for a circle if we approximate it by regular polygons. A circle of radius 1 has circumference [imath]2 \pi[/imath], and expected shadow size of 2 (since every shadow has the length 2), giving us the proportionality constant [imath]\pi[/imath].
This proof extends to any number of dimensions. So, the surface area of a 3D polyhedron is proportional to the average area of its shadow in a random direction onto a perpendicular plane, and the constant of proportionality can be found by considering the sphere (it's 4). Here, the average is taken over projection directions corresponding to the vectors on the unit sphere.