Thanks to several comments by Gerry Myerson, it is now clear that I wasn't clear, up to a state where I seriously confused myself. In a renewed attempt:
Recently, I've been thinking about Platonic solids (in preparation of a maths camp next summer) and about proofs that there are only five of these. There is one step in what seems to be the main proof that causes problems.
To avoid problems, polyhedra may contain infinitely many faces and Platonic solids consist of finitely many faces, all of these the same regular polygon, such that at every vertex the same number of polygons meet.
The proof I'm talking about considers the possibilities for $k$ regular $n$-gons meeting at each vertex ($k\geq3$) and considers when the sum of angles at each vertex (i.e. $k$ times the angle in a regular $n$-gon) is strictly less than $360^\circ$. This yields exactly five options and all five options can then be realised.
From considering counterexamples to several subtly wrong definitions of Platonic solids (in this setting), it seems that the requirement that $k$ times the angle in a regular $n$-gon) is strictly less than $360^\circ$ corresponds to the finiteness of the number of faces. I, however, fail to see why the finiteness of the number of faces implies this. (the implication in the other direction would be equally unclear, but irrelevant). Does anyone have suggestions how to (easily) prove that the finiteness of the number of faces implies that sum of angles at each vertex has to be strictly less than $360^\circ$? I still guess there's some obvious thing I'm missing.