Prove that a finite complete graph can be embedded in $\mathbb{R}^3$

Question

I've actually found a few intuitive examples where edges are taken to a twisted cubic and and the vertices are arranged in a certain way and that's very nice, but I'm actually more interested in a more straightforward algebraic proof if one exists. For example, showing a finite graph is also a noetherian topological space and then going from there (I'm not even sure that's true). I also came across a proof that used the fact that a finite graph is separable metric space. Basically I would just love to relate this problem in some way to abstract algebra if doable.