Ok, thank you Misha for the comments, let me try to fill out the hints you gave myself. I try to prove the following:
Let $g_n$ be a sequence of complete smooth metrics that converge in $C^0$ against the continuous metric $g$. Let $p$, $q \in M$ such that they are not cutpoints of each other w.r.t. any $g_n$. Let $\gamma_n$ be a $g_n$-geodesic connecting $p$ and $q$, then $\gamma_n$ converges to a $g$-geodesic in $C^0$.
By theorem 4.5 in the paper quoted above, for each $\eta>0$, there exists a number $N \in \mathbb{N}$ such that $$ (1-\eta)d_n(p, q) \leq d(p, q) \leq (1+\eta) d_n(p, q) ~~~~ \forall n\geq N$$ Hence for $\delta = \varepsilon/(1+\eta)$, $|s-t| < \delta$ implies $$ \varepsilon > (1+\eta)|s-t| = (1+\eta)d_n(\gamma_n(s), \gamma_n(t)) \geq d(\gamma_n(s), \gamma_n(t))$$ Hence the set of $\gamma_n$ with $n \geq N$ is equicontinuous and contained in some compact set by the same reasining as in the proof of the quoted thm. 4.5. Hence it subconverges in $C^0$ to a path $\gamma$. By the assumption that $p$ and $q$ are not cutpoints for any $g_n$, $\gamma_n$ is uniquely determined, so $(\gamma_n)$ converges itself (they fulfill a differential equation depending smoothly on the metric and hence are $C^0$ close if the metrics are $C^0$ close).
I do not yet see why $\gamma$ is a $g$-geodesic though.