Comment by shiandow

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.