The base case is n=2. The line joining them passes through exactly two points because that is all you have.
Now we attempt the induction step. We have n+1 points. Leave one, p, out. We know that the theorem applies to the n points by the induction hypothesis. So we have points q and r that have a line going through them. And the point of the theorem is that the line goes through only q and r, exactly two points of the n. All we have to do is add in p, not on that line, and we are done.
But we are also stuck. Point p is not one of the n points participating in the induction hypothesis. Nothing tells us that p is not on the line joining q and r.
So how do we prove it? It is a good, intriguing puzzle, but in proof theory, not geometry.
The Kochen-Specker theorem is interesting (if not useful). The proofs have a similar flavour to Sylvester. One tries to make a set of projections 'compatible' and it turns out to be impossible. Maybe there's a deeper connection.