It is only infinite for an infinitesimaly short coordinate length, so the integral is of course finite:
$\int_{r_{\rm s}}^{r_{0}} \sqrt{|g_{rr}|} \, dr =\int_{r_{\rm s}}^{r_{0}} \frac{1}{\sqrt{1-\frac{r_{\rm s}}{r}}} \, dr = \sqrt{(r_{0}-r_{\rm s}) r_{0}}+\ln \left(r_{0}+\sqrt{(r_{0}-r_{\rm s}) r_{0}}-\frac{r_{\rm s}}{2}\right) = {\rm finite}$
For the stationary bookkeeper this is still larger than $r_{0}-r_{\rm s}$, but smaller than $\infty$.
If you are in freefall with the negative escape velocity the gravitational depth expansion also exactly cancels out with the kinematic length contraction since $v_{\rm esc}=c \sqrt{r_{\rm s}/r}$, therefore $|g_{rr}|$ in raindrop coordinates is $1$, and the proper distance becomes exactly the coordinate distance.