 Solution

Use the standard meaning of static spacetime: a stationary spacetime chart whose time-space components also vanish. The printed static definition needs to inherit the timelike requirement $g_{00}<0$; without it, the requested timelike conclusion does not follow.

The preceding argument shows that $V={\partial }_0$ is a timelike Killing vector field. Its metric-dual one-form is $v=V_{\mu }d{x}^{\mu }=g_{00}d{x}^0$. Its exterior derivative satisfiesFor a torsion-free connection, antisymmetrizing the covariant derivative of a one-form gives its exterior derivative. Consequently the Killing vector has hypersurface orthogonality:Geometrically, the hypersurfaces $x^0=\mathrm{constant}$ are orthogonal to $V$.