Well, it certainly can't be done in a plane. I've seen some tricks to do it involving folding the paper as you draw the line and drawing on the back of the paper. Of course, that's not really helpful mathematically ;)

The proof I was taught involved the T junctions of the figure. Notice that whenever three segments meet in a T intersection, then when you are drawing the figure, one segment must begin or end at that intersection. If you do not start drawing at that point, then when you visit it for the first time, you obviously have to leave that point through a different segment. Thus, you have drawn two of the three segments, and there is only one segment remaining connected to that vertex. Obviously, when you draw that segment, there will be no way to leave that vertex again. For these kinds of problems, it follows that for every intersection of an odd number of segments, you must begin or end drawing a line there. Since the line you are drawing begins and ends somewhere, it can handle only two of the intersections with an odd number of segments. The figure you give has 8 T intersections, so it will take at least four seperate lines to draw.

This method can certainly disprove whether you can draw a figure in so many seperate, non-overlapping lines, but I am not sure if it can give the number of lines needed. It gives 4 lines as a floor, but I am not sure if it proves that you can actually draw the figure in 4 lines. It seems like it should, so I wonder if any figure can be drawn in so many lines as half the number of intersections with an odd number of segments. Does anyone know any proofs or counterexamples?

dave