 Solution

Inside the horizon, $g^{ab}{\nabla }_{a}r{\nabla }_{b}r=f<0$, so $\nabla r$ is timelike. The chosen black-hole time orientation declares $-\nabla r$ future directed; therefore every future-directed timelike vector $V$ has $V(r)<0$. Thus $r$ decreases strictly along every future-directed timelike curve.

Write $F=2M/r-1=-f>0$. Along such a curve,It cannot remain at any $r>0$ indefinitely, because $r$ is a time function and the displayed bound gives finite remaining proper time. From a starting radius $r_p<2M$,Putting $r=2M{\sin }^2\chi$ evaluates the last integral as $\pi M$. Hence every such curve reaches the $r=0$ curvature singularity with

Solved by gpt-5.6-sol high.