= Solution
For every member of the <geodesic variation>, <proper time> normalization gives $g(T,T)=-1$. By <metric compatibility>, the <geodesic equation>, and $\nabla_TS=\nabla_ST$,
$$
\frac{d}{d\tau}g(S,T)=g(\nabla_TS,T)=g(\nabla_ST,T)=\frac12S\bigl(g(T,T)\bigr)=0.
$$
The last equality uses normalization across the whole variation, not merely constancy along one curve. Hence $g(S,T)$ is constant along each <timelike geodesic>, proving
$$
\boxed{g(S,T)|_{\tau_0}=0\quad\Longrightarrow\quad g(S,T)=0\text{ for all }\tau.}
$$
This is <orthogonality of proper-time deviation vectors>.
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