[sweezyjeezy]

It's not it being case by case that's my issue. I used do olympiads and e.g. for the k>=3 case I wouldn't write much more than:

"Since there are 3k - 3 points on the perimeter of the triangle to be covered, and any sunny line can pass through at most two of them, it follows that 3k − 3 ≤ 2k, i.e. k ≤ 3."

Gemini writes:

Let Tk be the convex hull of Pk. Tk is the triangle with vertices V1 = (1, 1), V2 = (1, k), V3 = (k, 1). The edges of Tk lie on the lines x = 1 (V), y = 1 (H), and x + y = k + 1 (D). These lines are shady.

Let Bk be the set of points in Pk lying on the boundary of Tk. Each edge contains k points. Since the vertices are distinct (as k ≥ 2), the total number of points on the boundary is |Bk| = 3k − 3.

Suppose Pk is covered by k sunny lines Lk. These lines must cover Bk. Let L ∈ Lk. Since L is sunny, it does not coincide with the lines containing the edges of Tk. A line that does not contain an edge of a convex polygon intersects the boundary of the polygon at most at two points. Thus, |L ∩ Bk| ≤ 2. The total coverage of Bk by Lk is at most 2k. We must have |Bk| ≤ 2k. 3k − 3 ≤ 2k, which implies k ≤ 3.

[shiandow]

I'll admit I didn't look to deeply if it could be done simpler, but surely that is still miles better than what OpenAI did? At least Gemini's can be simplified. OpenAI labels all points and then enumerates all the lines that go through them.