Given $2n$ points in the plane, prove we can connect them with $n$ nonintersecting segments

This approach can work. If all the points would have different $y$ coordinates it is easy to separate the bottom two by a horizontal line. But if some points have the same $y$ coordinate we can just rotate the points around the origin by some angle.

Since there is only a finite number of pairs of points, and any given pair can have the same $y$ coordinate under only $2$ different angles of rotation. So there must be some angle such that all points end up with different $y$ coordinates.

Your argument works fine in an inductive proof, under the hypothesis of no collinear triples. For $n>1$ the convex hull of you points is a convex polygon that contains all the points, and any side of this polygon contains exactly two points (vertices of the polygon), and is disjoint from the convex hull of the remaining points. Those two points can be joined by a segment, and the disjointness condition guarantees that any segments provided by the induction hypothesis for the remaining points will not cross that segment.

I think it could even be adapted to work without the non-collinear hypothesis (as long as the points are distinct), though you need to be a bit more careful since one cannot necessarily pair two vertices of the convex hull; rather an appropriate pair of points on the boundary must be chosen.