First of all, the Peter-Weyl Theorem is about unitary representations, it will be of no use if you want to study representations on topological vector spaces which are not isomorphic to Hilbert spaces (or pre-Hilbert spaces). Thus, formally speaking, the answer to your question (at the end of the 1st paragraph) is negative. Nevertheless, one has:
Theorem. Let $V$ be a Hausdorff locally convex quasicomplete topological vector space, $G$ a compact (and Hausdorff) topological group and $\rho: G\to Aut(V)$ a continuous irreducible representation (meaning that $V$ contains no proper closed invariant subspaces). Then $V$ is finite-dimensional.
You can find a proof in
R.A.Johnson, Representations of compact groups on topological vectors spaces: some remarks, Proceedings of AMS, Vol. 61, 1976.
The point of considering this class of topological vector spaces (which includes, for instance, all Banach spaces) is that one has a satisfactory theory of integration for maps to such spaces. As for more general vector spaces, I have no idea, I suppose that the finite-dimensionality claim is simply false. For instance, if you drop the assumption that $V$ is Hausdorff (which one usually assumes) and take a vector space with trivial topology, you will have irreducible representations of any group on such a vector space.
One last thing: In the context of Hilbert spaces, there is a very short and direct (avoiding PWT) proof of finite dimensionality, given in this mathoverflow post.
Best Answer
Any nonzero vector $v\in V^G$ spans a $1$-dimensional trivial sub-representation. If $V$ is irreducible of dimension $>1$, then it follows that $V^G=0$. If $\dim V=1$, then $V^G\neq 0$ implies $V$ is the trivial representation.