One can assume that all line segments are actually lines, or half-lines, since you can let the lengths of the line segment go to infinity. Note that one infinite line segment creates two disjoint (convex!) regions. Any additional segment bisects a region, so you'll end up with $n+1$ convex regions. With $n+2$ points, pigeon hole principle finishes the proof.

One can assume that all line segments are actually lines, or half-lines, since you can let the lengths of the line segment go to infinity. Note that one infinite line segment creates two disjoint (convex!) regions. Any additional segment bisects a region, so you'll end up with $n+1$ convex regions. With $n+2$ points, pigeon hole principle finishes the proof.

EDIT: To address non-crossing, this is the detailed construction: Enumerate the intervals, and extend each interval, one by one, in both directions until they hit another interval (or to infinity). It does not matter if the some intervals are parallel or not. The non-overlapping condition ensures that the number of regions is exactly $n+1$ (or less, it two intervals determine the same line). Joseph's illustration is exactly this construction.

I like this problem, perhaps I'll 'steal' it to a collection of nice math problems.

The argument can be generalized to the analogous $n$ non-degenerate (full-dimensional) $n$-gons in $R^n$, and $n+2$ points. E.g., triangles in $R^3$, and $n+2$ points.