I just ran across this question, and thought I would give a precise version of the proof Ilya suggested. I believe I learned this proof in Richie Miller's topology course, Michigan State University, 1977 or so.
Choose a triangulation of the surface S, equipped with the simplicial metric. Choose a maximal one-ended subtree T of the dual 1-skeleton S(1). The subtree T contains every dual 0-cell, that is, the barycenter of every 2-simplex. Also, T contains dual 1-cells crossing certain 1-simplices. Let U be the union of the open 2-simplices and open 1-simplices that contain a point of T. The metric completion of U, denoted barU, is a closed disc with one boundary point removed, and so there is a deformation retraction from barU onto its boundary partialbarU. Attaching barU to S−U in the obvious way to form the surface S, the deformation retraction barUtopartialbarU induces a deformation retraction of S onto S−U, wnich is a subcomplex of the 1-skeleton.
By the way, the subtree TsubsetS(1) can be constructed by an explicit process. Enumerate the dual 0-cells v1,v2,ldotsinS(1). Construct one-ended subtrees T1,T2,ldotssubsetS(1) as follows. T1 is any proper ray based at v1. If vninTn−1 then Tn=Tn−1. If vnnotinTn−1, let Tn be the union of Tn−1 with any arc alphasubsetS(1) having one endpoint at vn and intersecting Tn−1 in its opposite endpoint. Each Tn is a one-ended tree by induction, and since the radius r neighborhood of v1 in Tn stabilizes as ntoinfty, it follows that T=cupnTn is a one-ended subtree of S(1), and it is maximal because it contains each vi.
I think this proof generalizes to any dimension, to give the theorem that Igor Belegradek refers to.
--- Edited to simplify and clarify the argument ---
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