TITLE: Visualization of surface area of a sphere
QUESTION [17 upvotes]: I help mentor some really young, bright kids in mathematics. We were looking at geometric properties of various shapes, and one of the kids noted that the surface area of a sphere $S = 4\pi r^2$ contains the equation for the area of a circle $A = \pi r^2$.
She was a bit confused why the factor of $4$ was mysteriously there. I told her I'd get back to her.
I know how to prove the formula using calculus, but I spent a long time trying to find an elementary way of doing it.
Does anyone know of a way of proving the first equation using almost no advanced mathematics$^1$? This seems unlikely, so as a separate question, does anyone know of a good visualization to show the relation between $S$ and $A$? 
The naive approach of taking four circles and showing you can "place them" on a sphere is clearly wrong (you can't just place four circles on a sphere), but I'm not sure what the alternative is.
$^1$These kids have a working knowledge of variable manipulation, basic geometry, and I guess combinatorics?

REPLY [2 votes]: I just got done teaching the class again and it went extremely well, so I wanted to post my methods here. This isn't technically an answer, but more of what I did which works as an answer.
All credit for my fundamental idea goes to @dxiv and the link he posted in the comment to my question.

I gave each kid half of a cored orange (essentially a hemisphere) and told them to use it as a stencil to trace out a circle. They determined the radius of this first circle to be $1.6 \text{ inches}$.
I argued that the surface area is a measure of how much paint we'd need to cover the top of the half-orange. I then argued that if we smushed the orange into a flat disk, the amount of paint we need obviously wouldn't change (if we painted every point pre-smush, no point would be unpainted post-smush), so the surface area wouldn't change.
They smushed the orange and, again, used it as a stencil to trace out a circle. They found the radius of this circle to be $2.25 \text{ inches}$.
The area of our original circle, $\pi r^2$, is about $8 \text{ inches}^2$. The area of our new circle is about $16 \text{ inches}^2$. So, the surface area of the hemisphere is $2\pi r^2$, and the surface area of the full sphere is $4\pi r^2$.
They were extremely happy with this demonstration, but I obviously wasn't - it wasn't quite rigorous. So to round out the lesson, I showed them the following diagram (reproduced here in Paint):

Under the smush operation, I said that the distance between the red and blue points should not change. This would mean that our second circle should have a radius that is off by a factor of $\sqrt2$ from our first circle. Lo and behold:
$$1.6\sqrt2 \approx 2.25$$
They all left pretty satisfied, so I think the demonstration worked, sans the fact it was a bit hand-wavy.