For an irreducible affine variety $X$, the dimension is defined to be the length of longest chain of prime ideals in the coordinate ring of $X$, $k[X]$.
Let $G^0$ denote the irreducible/connected component of an algebraic group $G$ containing $1$.
Assuming $\dim(G)= \dim(G^0)=0$, I have found online the following reasoning for why $G$ must be finite: a connected space of zero dimension is a single point, so that $G^0 = \{1\}$. Then because $G$ is the finite union of cosets of $G^0$, it follows that $G$ is finite.
Question: Why is a connected space of zero dimension just a point?
For the converse, I have only the idea that irreducible subvarieties of $X$ correspond to prime ideals in $k[X]$ by the Nullstellensatz. I'm not sure how this should give $\dim(G^0) = \dim(G) = 0$. Maybe because $G^0$ has no proper subvarieties so the conclusion follows?
Consider the coordinate ring of $X$. If $X$ is of dimension zero, there cannot be any chains of prime ideals with a nontrivial inclusion, so every prime ideal is maximal, which means that the coordinate ring is Artinian. Artinian rings are finite products of Artinian local rings, which means their Spectrum is a finite collection of points. All these statements are equivalences, so you can just run this argument in reverse to get the other direction.
Your idea that $G^0$ has no proper subvarieties is also correct- a chain of proper inclusions of prime ideals corresponds to a chain of proper inclusions of irreducible subvarieties, so no proper irreducible subvarieties implies dimension zero.