It can help to think about theorems like this by restating them as a puzzle asking for a counterexample.
Given N points, N > 2, can you arrange them in a Euclidean plane so that (1) they are not all on the same line, and (2) every line that goes through two of the points must also go through at least one more of the points?
Well put. A similar way to say it: ask the 9 year old to draw ALL of the possible lines connecting two points in the set — of course this is possible, just might take awhile. The theorem says that at least one of the lines drawn must hit ONLY the two points the 9 year old used when drawing that particular line, not any others.
I just thought of another way to restate. Suppose you give me any finite set of points, any finite set at all, and you also tell me that when the 9 year old draws any line through any two of them, she will always hit a third. Then I can immediately conclude all the points lie on a single line, that is, that any line the 9 year old draws will hit all the points.
Given N points, N > 2, can you arrange them in a Euclidean plane so that (1) they are not all on the same line, and (2) every line that goes through two of the points must also go through at least one more of the points?
The theorem says that you cannot do this.