= Solution
In first-order <time-dependent perturbation theory>, the resonant time factor before taking the <Fermi's golden rule> limit is
$$
\left|\int_0^t e^{i\nu t'}\,dt'\right|^2
=\frac{4\sin^2(\nu t/2)}{\nu^2}.
$$
Its central frequency width is of order $1/t$, and its total area is $2\pi t$. To replace it by $2\pi t\delta(\nu)$, the oscillator <density of states> and relevant occupations must be approximately constant over that width. Thus $t\Delta\omega\gg1$. At the same time, the perturbative flip <probability> must remain much smaller than one.
Define $\Gamma_0=2\pi\lambda^2g(\Omega)/\hbar^2$, so the two flip rates are $\Gamma_0(n_\Omega+1)$ and $\Gamma_0n_\Omega$. A sufficient common time window for both initial spins is
$$
\boxed{\frac1{\Delta\omega}\ll t\ll
\frac1{\Gamma_0(n_\Omega+1)}.}
$$
For a specified initial spin, the upper condition is simply its own flip rate times $t\ll1$. Such a window requires weak enough coupling. The continuum approximation also presumes times shorter than recurrences from a resolved finite oscillator spectrum, and long enough to resolve the positive-frequency resonance from its nonresonant counterpart.
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