We define a projective plane of order $q$ as a $(q+1)$-uniform family of subsets of a set $X$ where $|X|=q^2+q+1$ such that there is exactly one line passes every two distinct points.
Here we are given with a subset $S$ of $X$ of size $|S|=q+2$ such that none of any three points in $S$ is on a line. We need to show that for every line $L$, if intersects $S$, then $|L\cap S|=2$.
By this assumption, it is clear that if $L$ intersects $S$, then $1\leq|L\cap S|\leq 2$. But I just cannot come up with a method to exclude the possibility $|L\cap S|=1$, even though I know it is wrong. I am wondering on how we can obtain a contradiction if we assume $|L\cap S|=1$.
Consider a point $P\in S$. There are $q+1$ lines through $P$. None of them meets $S$ in more than one point other than $P$. As there are $q+1$ further points in $S$ then each of the $q+1$ lines through meets $S$ in one point other than $P$, that is $|L\cap S|=2$.
Such sets $S$ are hyperovals.