Sylvester–Gallai theorem (Kelly's minimum-distance proof)

Step 1 (set up the extremal choice). Let $P$ be the given finite set of $n \ge 3$ points, not all collinear. Consider the finite set of pairs $(Q, \ell)$ where $\ell$ is a line through at least $2$ points of $P$ and $Q \in P$ is a point not on $\ell$. This set is nonempty (since $P$ is not all collinear) and finite, so by the extremal principle we may choose a pair $(Q_0, \ell_0)$ minimizing the distance $d(Q_0, \ell_0)$ from the point to the line.

Step 2 (assume a contradiction). Suppose, for contradiction, that $\ell_0$ contains at least $3$ points of $P$. Let $F$ be the foot of the perpendicular from $Q_0$ to $\ell_0$. Since there are $\ge 3$ points of $P$ on $\ell_0$ and they lie on at most $2$ rays emanating from $F$ along $\ell_0$, the pigeonhole principle gives two of them, $B$ and $C$, on the same ray, with $B$ between $F$ and $C$ (allowing $B = F$).

Step 3 (build a closer pair via similar triangles). Drop a perpendicular from $B$ to the line $Q_0C$, with foot $G$. The right triangles $\triangle BGC$ and $\triangle Q_0FC$ share the angle at $C$, so they are similar, giving $\dfrac{BG}{Q_0F} = \dfrac{BC}{Q_0C}$. Since $B$ lies between $F$ and $C$ we have $BC \le FC$, and since $\triangle Q_0FC$ has a right angle at $F$, the hypotenuse satisfies $FC < Q_0C$; combining these, $BC < Q_0C$, hence $BG < Q_0F = d(Q_0, \ell_0)$.

Step 4 (contradiction). The line $Q_0C$ passes through $2$ points of $P$ (namely $Q_0$ and $C$), and $B$ is a point of $P$ not on it, so $(B, Q_0C)$ is a valid pair in our finite set with $d(B, Q_0C) = BG < d(Q_0, \ell_0)$, contradicting the minimality of $(Q_0, \ell_0)$.

Step 5 (conclusion). The contradiction shows $\ell_0$ cannot contain $3$ or more points of $P$; since it was chosen to contain at least $2$, it contains exactly $2$, so $\ell_0$ is the required ordinary line.