My book (Isham's Lectures on Group Theory and Vector Spaces for Physicists) claims
Every unitary representation $g \to U(g)$ on a Hilbert space $\mathcal H$ of any group $G$, is completely reducible.
The first line of the proof is
Let $W$ be any $G$-invariant subspace of $G$.
How is such a subspace known to exist? I am not familiar enough with Hilbert space theory, but presumably there is a generalization to the finite-dimensional result that every unitary operator on a complex space is diagonalizable. But why should there in general be an eigenspace which is an eigenspace for each $U(g)$?
An invariant subspace always exists: $\{0\}$ is one an the whole Hilbert space is another. If there is no other (closed) invariant subspace, the representation is said to be irreducible.
I don’t have the book at hand, but, for finite-dimensional Hilbert spaces, at least, the proof goes like this: if the representation is irreducible, we’re done; if not, one can decompose it as the direct sum of two closed invariant subspaces, and we try to decompose these… until we’re done.
However, for infinite-dimensional Hilbert spaces and some non-compact groups, one cannot, in general, decompose a representation as a direct sum of irreducible subrepresentations. We have to generalize the concept of direct sum to the one of direct integral. But this is another story.