For points inside the event horizon, with the exception of the singularity itself, it's actually very easy to see without doing any intensive math. For any spacelike curve $$x^{\mu}(\lambda)$$ expressed in terms of some parameter $$\lambda$$ starting at $$\lambda_{0}$$ and ending at $$\lambda_{1}$$, the spacetime distance is just the integral
$$s \,=\, \int_{\lambda = \lambda_{0}}^{\lambda = \lambda_{1}} \mathrm{d}s \,=\, \int_{\lambda_{0}}^{\lambda_{1}} \mathrm{d} \lambda \sqrt{g_{\mu\nu} \frac{\mathrm{d}x^{\mu}}{\mathrm{d}\lambda} \frac{\mathrm{d}x^{\nu}}{\mathrm{d} \lambda}} \,.$$
A simple example you could consider is just to draw a horizontal line on a
Kruskal diagram starting at some Kruskal (spacelike) coordinate $$u_{0}$$ inside the event horizon and ending at some Kruskal coordinate $$u_{1}$$ outside the event horizon, at some fixed Kruskal time coordinate $$v \,>\, 0$$. If you use the Kruskal coordinate $$u$$ itself as the line parameter, then $$\frac{\mathrm{d}v}{\mathrm{d}u} \,=\, 0$$ and $$\frac{\mathrm{d}u}{\mathrm{d}u} \,=\, 1$$, and the spacetime distance just becomes
$$
\begin{eqnarray}
s &=& \int_{u_{0}}^{u_{1}} \mathrm{d}u \sqrt{g_{vv} \Bigl( \frac{\mathrm{d}v}{\mathrm{d}u} \Bigr)^{2} \,+\, g_{uu} \Bigl( \frac{\mathrm{d}u}{\mathrm{d}u} \Bigr)^{2}} \\
&=& \int_{u_{0}}^{u_{1}} \mathrm{d}u \sqrt{g_{uu}} \,.
\end{eqnarray}$$
At this point, you could crack open a textbook or consult Wikipedia to find out exactly what $$g_{uu}$$ is for the black hole metric in Kruskal coordinates (NB: it's a function of $$v$$ and $$u$$), substitute it in, and try to calculate the integral to get an exact value. In the case of an eternal black hole, Wikipedia says it's
$$g_{uu} \,=\, \frac{32 G^{3} M^{3}}{r} e^{- r / 2GM}$$
with $$r \,=\, r(v,\, u)$$ defined implicitly by $$v^{2} \,-\, u^{2} \,=\, ( 1 \,-\, r/2GM ) e^{r/2GM}$$. But for our purposes this is more detail than we need. The only thing you need to notice about $$g_{uu}$$ is that it is finite everywhere except on the singularity (where $$r \,=\, 0$$). Consequently, the spacetime interval is the integral of a finite function over a finite domain (unless you somehow managed to draw an infinitely long line on a piece of paper, $$u_{1} \,-\, u_{0}$$ is finite), which will obviously produce a finite value.
