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A projective plane of order $q$ has $q^2+q+1$ points and lines, with $q+1$ lines through each point, and exactly one line through two distinct points. For a set $S$ of size $s$, put $x_\ell=|S\cap\ell|$. Double counting gives

$$\sum_\ell x_\ell=s(q+1),\qquad \sum_\ell x_\ell(x_\ell-1)=s(s-1).$$

If $1\le x_\ell\le C$, then $x_\ell(x_\ell-1)\le C(x_\ell-1)$. Summing and rearranging proves

$$s^2-[C(q+1)+1]s+C(q^2+q+1)\le0.$$

Any cap at most three also satisfies the hypotheses with $C=3$. In that case the discriminant is $-3q^2+12q+4$, which is negative at $q=5$ and strictly decreasing for real $q\ge5$. The monic quadratic is positive for every real $s$, a contradiction. The plane over the field of five elements exists, so a cap working for every finite plane cannot be less than four: a real bound below four permits at most three points on a line.

For $C=4$ the discriminant is $24q+9$, giving the interval on the claim page. A necessary interval does not supply a point set attaining it.