Proof that two non-parallel planes must intersect? I managed to find, by enumeration, the intersection point of two planes $ax+by+cz+d=0$ and $ex+fy+gz+h=0$, in all possible cases (with the condition that the planes are not parallel). But this is a very ugly proof.
I wonder if there is a quicker and more elegant proof (without linear algebra --- this is high school level (Euclidean) geometry)?

"I managed to find, by enumeration, the intersection point of two planes ax+by+cz+d=0 and ex+fy+gz+h=0, in all possible cases (with the condition that the planes are not parallel). But this is a very ugly proof."
THis is not a proof as analytic geometry was defined with the Euclid's 5th postulate incorporated into the entire basis of converting lines and planes to coordinate system. so this is circular.
"I wonder if there is a quicker and more elegant proof (without linear algebra --- this is high school level geometry)?
No.  Because it isn't true.  Read up on non-euclidean geometry.
Euclid's 5th postulate-- that if the interior angles of two line and an mutually intersecting third line are supplimentary than the lines never intersect, but if not, the lines will eventually intersect on the side where the interior angles are less than supplementary--- is an axiom.  
If we modify the axiom to render that there are infinite number of lines that will not intersect with interior angles of varying measurements, or if we modify it so that all pairs of lines will always intersect twice even if the interior angles are supplimentary--- then the new axiom will yield different but equally valid systems of geometry.
