Prove that a set of lines, each of which intersect the others, are either coplanar or share a single common point.
Source: Hadamard
This is a surprisingly difficult problem, because it requires ping-ponging back and forth between two dissimilar concepts: lines being coplanar and lines sharing a common point.
I could not find a way to unify these two concepts, so instead developed the proof below, to which I request verification and feedback. Is the proof correct? Is it well written (I tried to provide sufficient detail while being succinct)? Is there an alternate or simpler solution?
Proof: Let $S$ be a set of lines all of which share a common point $A$. If line $\ell$ does not include point $A$, but still intersects every line in $S$, then $S \cup \{\ell\}$ must be coplanar. Indeed, there is a unique plane $P$ containing $\ell$ and $A$, and since every line in $S$ includes two points on $P$ (the line's point of intersection with $\ell$ and the point $A$), every line in $S$ lies within $P$.
Furthermore, if $S$ is a set of coplanar lines, all lying in plane $P$, that do not share a [edit: delete single] common point, and $\ell$ intersects every line in $S$, then $\ell$ must also lie in $P$. Indeed, there must be at least two distinct points where $\ell$ intersects a line from $S$, and since both these points are on $\ell$ and in plane $P$, then $\ell$ lies in $P$.
The claim therefore follows by induction, starting from the trivial case of [edit:] any two lines [previously was: the empty set].
As requested, below is the original problem from Hadamard, in the original French, taken from Leçons De Géométrie Élémentaire, Volume 2, Livre V, Page 6, Problème 423:
First notice that "either or" is false, it should be an inclusive or: a set $S$ of lines which share a single common point can be coplanar (and even must be, if $|S|=2$).
Your proof looks fine now, but here is a variant without induction:
Let $S$ be your set of (at least two) lines, each of which intersects the others. Let $\ell_1,\ell_2$ be two of them, sharing a point $A$ and lying in a plane $P$.
Assume that not all lines of $S$ contain $A$, and let us prove that all lines of $S$ lie in $P$. Let $\ell_3$ be a line in $S$ which does not contain $A$. Then, the three lines $\ell_1,\ell_2,\ell_3$ don't share a common point. In particular, $\ell_3$ lies in $P$ since it contains two distinct points of $P$: one in $\ell_1$ and one in $\ell_2$. Therefore, for every $\ell\in S$, $\ell$ meets their union $\ell_1\cup\ell_2\cup\ell_3$ in at least two distinct points. Again, since these points lie in $P$, so does $\ell$.