= Solution
Pass to $B=\overline A^{\|\cdot\|}$. Closed invariant subspaces and intertwining maps are unchanged by this closure, and $BH$ is still dense. Because $B$ 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 $b$; an isolated nonzero <eigenvalue> gives a nonzero finite-rank spectral projection $e\in B$ by <continuous functional calculus>, using a function vanishing at zero. The finite-dimensional corner $eBe$ has a minimal nonzero projection $p$, with $pBp=\mathbb Cp$. Choose a unit vector $v\in pH$ and set $K=\overline{Bv}$. Then $pK=\mathbb Cv$, since $pbv=pbpv$ is scalar times $v$. If $Q$ is the projection onto a reducing subspace of $K$, it commutes with $B$, and $Qv=pQv\in\mathbb Cv$. Its projection property makes $Qv$ either zero or $v$. Cyclicity then makes $Q=0$ or $I_K$. Thus $K$ 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 $B$ would, by taking adjoints, be orthogonal to $BH$ and hence zero. The image of $B$ 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 $H$.
Finally, suppose one irreducible equivalence class occurred infinitely often. Choose some $b\in B$ acting nontrivially on that class, and a unit vector $v$ with $\|bv\|>0$. Transport $v$ by unitary intertwiners into distinct equivalent summands. These vectors are orthonormal; their images under $b$ are orthogonal and all have the same nonzero norm. That contradicts compactness of $b$. \b[The decomposition has finite multiplicity for every irreducible type], as claimed in <nondegenerate representations by compact operators>.
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