= Solution
\b[No: the lowest dissociation channel is $\mathrm{He}+\mathrm H^+$, which a doubly occupied helium-centred orbital can describe.] In the heteronuclear <molecular orbital> diagram the helium $1s$ site lies below the hydrogen $1s$ site. As separation grows, the bonding orbital tends to helium $1s$, the antibonding orbital tends to hydrogen $1s$, and the two ground-state electrons remain in the lower orbital.
There is therefore no symmetry requirement to keep both sites equally occupied and no compulsory half-weight mixture of wrong ionic configurations at dissociation. The physical asymptote is already a closed-shell atom plus a bare proton. The alternative $\mathrm{He}^++\mathrm H$ limit lies higher. A <Restricted Hartree-Fock method> curve approaches the Hartree–Fock helium energy rather than the exact correlated helium energy, so ordinary <dynamic electron correlation> is still missing, but this is not the homonuclear static-correlation catastrophe. In the preceding two-panel sketch, the helium hydride curves have the same qualitative dissociation channel with a small residual correlation offset, whereas the hydrogen restricted curve reaches a qualitatively wrong high limit.
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