Suppose that $S$ is the surface, $\alpha$ is the arc, and $K$ is the compact set. 

It suffices to find a sequence $(B_n)$ of two-cells which are nested neighbourhoods of $\alpha$, and whose intersection is $\alpha$.  So from now on we ignore the set $K$.  (*) We next find an arc $\beta$ so that $\alpha \cap \beta = \partial \alpha = \partial \beta$.  That is, the union of the two arcs is a Jordan curve.  We now apply some version of the Jordan curve theorem to find an annulus (or Möbius) neighbourhood of $\alpha \cup \beta$. We cut this down in various ways to get the first two-cell neighbourhood $B_0$.  All the other $B_n$ come similarly. 

Of course, (*) is asking if the endpoints of $\alpha$ are "accessible".  Showing such things is part of the machinery of various proofs of the Jordan curve theorem.  There may be some other way to package things; however, I would be very surprised if there was a proof that avoids using the JCT (and Schoenflies) or its parts...