In the following proof (after "But there is a more elementary proof"), I was confused on something.
Apparently we can assume without loss of generality that $V = V_{\chi}$. In this case, here is what I think Springer is saying: let $H = (G,G)$, and assume $G$ is acting as a group of automorphisms on $V$. Every $x \in H$, interpreted as a linear operator on $V$, has determinant one. At the same time, $x.v = \chi(x)v$ for all $x \in H$. Thus the matrix of transformation of any $x \in H$ is $$\begin{pmatrix} \chi(x) & & 0 \\ & \ddots & \\ 0& & \chi(x) \end{pmatrix}$$ with $\chi(x)^n = 1$. Thus $H$ is finite, because there are only finitely many such matrices (actually now that I think about it, I don't see why that's true either; $\chi$ could be the trivial character)).
My issue with this proof is that in the argument above, we have injected $H$ into a group of matrices, and showed that that group was finite. On the other hand, if we don't assume that $V = V_{\chi}$, then we would be taking elements of $x \in G$, interpreting those elements as automorphisms of $V$, and then restricting those automorphisms to the $G$-stable subspace $V_{\chi}$. In that case, the image of $H$ is finite, as Springer argued. How does that mean that $H$ itself is finite? We do not even know that $V$ is the direct sum of the $V_{\chi}$.