= Solution
Put $f(r)=1-2M/r$. An incoming radial null ray has $dt/dr=-1/f$. The <tortoise coordinate> $r_*=r+2M\log|r/(2M)-1|$ obeys $dr_*/dr=1/f$, so $v=t+r_*$ is constant along such a ray. For $r>2M$ this is exactly the printed logarithm. Substituting $dt=dv-dr/f$ gives the metric in <Ingoing Eddington-Finkelstein coordinates>
$$
\boxed{ds^2=-f\,dv^2+2\,dv\,dr+r^2d\Omega^2.}
$$
The radial null directions satisfy either $dv=0$ or $dr/dv=f/2$. Future-directed ingoing rays have decreasing $r$ on the first family. Outgoing rays increase $r$ outside the horizon, stay at $r=2M$ on its generators, and decrease $r$ inside. Thus inside the horizon every future timelike trajectory also moves toward smaller $r$.
The $(v,r)$ metric block has <determinant> minus one and is nonsingular at $r=2M$. The horizon is a null surface, not a <curvature singularity>. For example the <Kretschmann scalar> $48M^2/r^6$ is finite there and diverges at $r=0$.
Inside, set $A=2M/r-1>0$. For radial timelike motion, the Schwarzschild expression becomes $d\tau^2=A^{-1}dr^2-A\,dt^2$, hence
$$
\boxed{d\tau\le\sqrt{\frac r{2M-r}}\,|dr|.}
$$
Angular motion would only reduce the available <proper time> further. Integrating from the horizon to zero, with $r=2M\sin^2\theta$, gives
$$
\boxed{\Delta\tau\le\int_0^{2M}\sqrt{\frac r{2M-r}}\,dr=4M\int_0^{\pi/2}\sin^2\theta\,d\theta=\pi M.}
$$
The bound is sharp for the zero-energy radial <geodesic> with constant Schwarzschild $t$ in the interior. For a <geodesic> arriving from the exterior with strictly positive conserved energy, the <proper time> is strictly smaller; $\pi M$ is the supremum as that energy tends to zero. Thus the universally available horizon-to-singularity upper bound is $\pi M$, rather than a time attainable by every exterior infaller. Restoring units gives $\pi GM/c^3$.
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