The answer to the question as stated is clearly "no": take for $S$ the plane $\mathbb{C}$ with the metric induced by the differential $z^2\mathrm{d}z$; z\mathrm{d}z$; then$S^{\ast }=\mathbb{C}\smallsetminus {0}$and there is simply no geodesic from$1$and$\mathrm{i}$. Or, if you want closed loops and free homotopy, there is no geodesic in the free homotopy class of the loop around$0$. Intuitively, it is clear what happens: if you draw a curve and try to pull it straight, you are forced to go through the cone point at the origin. If, however, you consider the surface$S$itself and extend the definition of a geodesic so as to work on Euclidean metrics with cone points as well, the answer is "yes" if$S$is complete as a metric space. The usual definition of geodesics in this context is one which works for all metric spaces: locally isometric maps from an interval. The statement you asked for is then a consequence of the ArzelĂ -Ascoli theorem. These things are treated in detail in the textbook Quadratic differentials by Kurt Strebel. 1 The answer to the question as stated is clearly "no": take for$S$the plane$\mathbb{C}$with the metric induced by the differential$z^2\mathrm{d}z$; then$S^{\ast }=\mathbb{C}\smallsetminus {0}$and there is simply no geodesic from$1$and$\mathrm{i}$. Or, if you want closed loops and free homotopy, there is no geodesic in the free homotopy class of the loop around$0$. Intuitively, it is clear what happens: if you draw a curve and try to pull it straight, you are forced to go through the cone point at the origin. If, however, you consider the surface$S$itself and extend the definition of a geodesic so as to work on Euclidean metrics with cone points as well, the answer is "yes" if$S\$ is complete as a metric space. The usual definition of geodesics in this context is one which works for all metric spaces: locally isometric maps from an interval. The statement you asked for is then a consequence of the ArzelĂ -Ascoli theorem.