= Solution
A gauge-invariant observable is <BRST-closed>: replacing its infinitesimal gauge parameter by the ghost gives $s\mathcal O=0$. Change the gauge functional continuously, or interpolate between two admissible choices, by changing the <gauge-fixing fermion> to $\Psi_t$. The action changes by the <BRST-exact operator> $\partial_tS=s(\partial_t\Psi_t)$.
For a normalized correlator of $\mathcal O=\prod_i\mathcal O_i$ with each insertion <BRST-closed>, differentiation of the <functional integral> gives
$$
\partial_t\langle\mathcal O\rangle
=i\left[\langle\mathcal O\,s(\partial_t\Psi_t)\rangle
-\langle\mathcal O\rangle\langle s(\partial_t\Psi_t)\rangle\right].
$$
The <BRST Ward identity> says $\langle sX\rangle=0$ for an invariant measure and action, with appropriate boundary conditions. Since $s\mathcal O=0$, the <graded Leibniz rule> makes the first insertion an exact variation of $\mathcal O\,\partial_t\Psi_t$ up to its harmless parity sign, and both terms vanish. Thus
$$
\boxed{\partial_t\langle\mathcal O_1\cdots\mathcal O_n\rangle=0.}
$$
\b[Physical gauge-invariant correlation functions are independent of this gauge condition], even though individual gauge-field and ghost <Feynman diagrams> change. This argument assumes an admissible perturbative gauge fixing, a <BRST symmetry>-preserving regulator/measure and no uncanceled boundary contribution. A global failure of those assumptions is not settled by the formal local calculation.
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