 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 ${\partial }_{t}S=s({\partial }_t{\Psi }_t)$.

For a normalized correlator of $\mathcal{O}={\prod }_i{\mathcal{O}}_i$ with each insertion BRST-closed, differentiation of the functional integral givesThe BRST Ward identity says $⟨sX⟩=0$ 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. ThusPhysical 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.