What Emerton says is of course correct: an irreducible representation of a finite group is necessarily finite-dimensional. Indeed, the same holds for any continuous representation of a compact group on a Hilbert space.

It seems to me that you can get away with milder assumptions: $G$ can be any group so long as $V$ is a Banach space and $A$ is a bounded linear operator. Then in the proof you take $\alpha$ to be the Banach space analogue of an eigenvalue for $A$, i.e., an element of the spectrum of $A$.

You did of course look ahead and try to see what form of Schur's Lemma is actually used? (I mentioned a generalization because of the vague impression that operators discussed in physics are usually on infinite-dimensional spaces.) There are certainly multiple related results all going under that name. If you feel you need a different form than is proved in the text, let us know. I'm sure someone here (e.g. Emerton) can help you out.

What Emerton says is of course correct: an irreducible representation of a finite group is necessarily finite-dimensional. Indeed, the same holds for any continuous irreducible representation of a compact group on a Hilbert space.

It seems to me that you can get away with milder assumptions: $G$ can be any group so long as $V$ is a Banach space and $A$ is a bounded linear operator. Then in the proof you take $\alpha$ to be the Banach space analogue of an eigenvalue for $A$, i.e., an element of the spectrum of $A$.

You did of course look ahead and try to see what form of Schur's Lemma is actually used? (I mentioned a generalization because of the vague impression that operators discussed in physics are usually on infinite-dimensional spaces.) There are certainly multiple related results all going under that name. If you feel you need a different form than is proved in the text, let us know. I'm sure someone here (e.g. Emerton) can help you out.