The Sylvester-Gallai theorem

Choosing a minimal pair. There are finitely many such pairs $(P,\ell)$, and at least one. Choose one for which the distance $h=\operatorname{dist}(P,\ell)>0$ is smallest. Let $Q$ be the foot of the perpendicular from $P$ to $\ell$.

Claim. The line $\ell$ contains exactly two points of $S$.

Suppose not: $\ell$ contains at least three points of $S$. Among them, at least two lie in one closed half-line of $\ell$ starting at $Q$. Call these two points $B$ and $C$, with $B$ between $Q$ and $C$ or $B=Q$, and $B\ne C$. Then $BC\le QC<PC$.

A closer pair. Consider the line $m=PC$, which contains the two points $P,C$ of $S$. The point $B$ is not on $m$. The distance from $B$ to $m$ is
$$\operatorname{dist}(B,m)=BC\sin\angle PCQ=BC\cdot\frac{h}{PC}<h,$$
because $BC<PC$. So the pair $(B,m)$ has a smaller distance than $(P,\ell)$, contradicting the choice.

Therefore $\ell$ passes through exactly two points of $S$. $\square$

Remark. This is Kelly's proof (1948) of a conjecture of Sylvester. The statement is false in the projective plane (the Fano plane has $7$ points and every line has $3$), which shows that the proof must use distances.

Lesson. The extremal principle: pick the configuration minimising a quantity and show that a smaller one exists unless the structure is special.