@TedShifrin the argument is neat, I think. This is the statement: let $G$ be a profinite group and $\rho: G \to \operatorname{GL}_n(\Bbb C)$ be a continuous representation. Then the image of $\rho$ is finite. The proof of my lecturer was more elementary, but more involved, I did this:
the image is compact, so closed, thus a Lie subgroup, so in particular locally connected like all manifolds. But since the kernel is a closed subgroup of $G$, the quotient $G/\operatorname{ker}(\rho)=\operatorname{im}(f)$ is again profinite, so totally disconnected (note: continuous images of totally disconnec…

@TedShifrin the statement is quite remarkable.
It says that if we want to understand finite-dimensional complex representations of a complicated, but important group like $\operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q)$, we get them all by doing the "obvious thing" to construct representations: choose a finite Galois extension $K/ \Bbb Q$, then we have complex representation theory of finite groups, which is well understood and compose everything with the restriction map $\operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q) \to \operatorname{Gal}(K/\Bbb Q)$