A finite set $\mathfrak{S}$ of $n$ points in the Euclidean Plane has the property that any line through two of them passes through a third. Show that all the points lie on a line
Here's my proof using induction which I hope, is correct, though I somewhat believe there's a loophole in it.
So I was thinking if—
For $3$ points in the plane, it's obvious. For $4$ points $A,B,C,D$, choose $A,B$. A line through $A,B$ implies that $C$ or $D$ is a part of it since if they don't belong to $AB$ (extended) then the statement isn't valid. So, WLOG $C$ belongs to $AB$ (extended). Now, $A—B—C$ (in some order). Join $C,D$ and $A,C$. You get a $\triangle$ which has to be degenerate for the statement to hold.
Now for the induction step, assume there are $k$ ($k\geq 3$) collinear points in the plane. For every other point in the plane that doesn't belong to the line, we can draw a $\triangle$ which has to be degenerate for the statement to hold.
While the proof seems correct in essence, the triangle argument is slightly hand wavy as written. I would suggest you write it as follows:
Induction step: Consider points $P_1, P_2, ..., P_k$ for $k \geq 5$. By the induction hypothesis, the first $k-1$ points lie on the same line. However, since the induction hypothesis holds for any subset of these points, we can also state this for the last $k-1$ points. This means that points $P_2, P_3, ..., P_{k-1}$ all lie on the same line. Since $k-2 \geq 2$ there is only one such line, and therefore $P_1$ and $P_k$ also lie on that line.