 Solution

For a radial timelike geodesic, the static Killing energy is $E=f\dot{t}$ and normalization gives ${\dot{r}}^2=E^2-f$. Choose $E=\sqrt{2}$ and start at $r(0)=0$ with outward velocity one. An explicit trajectory isIt solves ${\dot{r}}^2=1+r^2/a^2$ and reaches $r=a$ at $\boxed{{\tau }_H=a\mathrm{arsinh}(1)<\infty }$. Differentiating the energy equation gives $\ddot{r}=r/a^2$, the radial geodesic equation, so this is an actual freely falling trajectory.

The regular coordinate satisfies $\dot{u}=(E-\dot{r})/f=1/(E+\dot{r})$, which is finite and positive at the crossing. Together with $u(\tau )=u(0)+{\int }_0^{\tau }[\sqrt{2}+\cosh (s/a){]}^{-1}ds$, the trajectory extends through the horizon in the regular chart. Divergence of the static time there does not imply infinite proper time.