Let $P=(S,L)$ be a projective plane of order $n$. Let $K$ be a nonempty subset of $S$ with the property that no three points belonging to $K$ are collinear. Prove that $|K|$ is less than or equal to $n+2$.
I began by staying that since $P$ is of order $n$, there are exactly $n^2+n+1$ points in $S$ and since $K$ has the stipulation that no three points are collinear, the points in $K$ must be at least on two lines. Now here is where I'm stuck. I don't know how to go from there to less than or equal to $n+2$. Any help would be appreciated.
Suppose $K$ is a set of $n+2$ points, no three of which are collinear. Let $x$ be any point which is not in $K$. Every point in $K$ is on some line passing through $x$. But there are only $n+1$ lines passing through $x$, so two points of $K$ must be on the same line through $x$. Thus $K\cup\{x\}$ does contain three collinear points. Since $x$ was an arbitrary point not in $K$, this shows that $K$ cannot be enlarged beyond $n+2$ points.