Pass to . Closed invariant subspaces and intertwining maps are unchanged by this closure, and is still dense. Because is a star-algebra, every closed invariant subspace reduces it.
Every nonzero nondegenerate compact-operator representation contains an irreducible reducing subspace. Indeed choose a nonzero positive compact element ; an isolated nonzero eigenvalue gives a nonzero finite-rank spectral projection by continuous functional calculus, using a function vanishing at zero. The finite-dimensional corner has a minimal nonzero projection , with . Choose a unit vector and set . Then , since is scalar times . If is the projection onto a reducing subspace of , it commutes with , and . Its projection property makes either zero or . Cyclicity then makes or . Thus is irreducible.
Take a maximal orthogonal family of such irreducible reducing subspaces, by the axiom of choice. If their orthogonal complement were nonzero, its restricted representation would be nondegenerate: a vector annihilated by all restricted would, by taking adjoints, be orthogonal to and hence zero. The image of on this complement is a norm-closed C-star algebra of compact operators, so the preceding construction gives another irreducible summand, contradicting maximality. The family therefore exhausts .
Finally, suppose one irreducible equivalence class occurred infinitely often. Choose some acting nontrivially on that class, and a unit vector with . Transport by unitary intertwiners into distinct equivalent summands. These vectors are orthonormal; their images under are orthogonal and all have the same nonzero norm. That contradicts compactness of . The decomposition has finite multiplicity for every irreducible type, as claimed in nondegenerate representations by compact operators.

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