Suppose that $S$ is the surface, $\alpha$ is the arc, and $K$ is the compact set.
It suffices to findProof 1: Find an arc $\alpha'$ so that $\gamma = \alpha \cup \alpha'$ is a sequenceJordan curve $(B_n)$ of two-cells which are nested neighbourhoods(+). The Jordan/Schoenflies [J/S] theorems provide an annulus (or Möbius) neighbourhood $A$ of $\alpha$,$\gamma$ and whose intersection isa homeomorphism on $\alpha$$A$ making $\gamma$ "straight". So from now on we ignore the set Shrink $A$ as needed to avoid $K$. (*) We next and then use $A$ to find an arcthe desired two-cell.
Proof 2: Let $\beta$ so that$T$ be the double branched cover of $\alpha \cap \beta = \partial \alpha = \partial \beta$. That is$S$, branched over the uniontwo points of $\partial \alpha$. Let $\beta$ and $\beta'$ be the two arcspreimages of $\alpha$ in $T$. So the union $\gamma = \beta \cup \beta'$ is a Jordan curve. We now apply some version of the Jordan curve theorem Apply [J/S] to find an annulus (or Möbius) neighbourhood $A$ of $\alpha \cup \beta$$\gamma$. We cut this down in various ways to get Shrinking $A$ as needed, we may assume that it is disjoint from the firstpreimage of $K$. We further arrange matters so that $A$ is invariant under the deck transformation, as are two-cell neighbourhood of its cutting arcs $B_0$(meeting $\gamma$ in exactly the points $\partial \beta$). AllSo the otherimage of $B_n$ come similarly$A$ is the desired two-cell.
Of course, in Proof 1, sentence (*+) is asking if thesaying that endpoints of $\alpha$ are "accessible". Showing such things is part of the machinery of various proofs of the Jordan curve theorem[J/S]. There may be some other wayThe branched cover trick in Proof 2 is designed to package things; however, I would be very surprised if there was a proof that avoids using the JCTavoid (and Schoenflies+). However, I'll guess that no proof can avoid using [J/S], or at least some of its parts.."working parts".