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 $\bar U$, is a closed disc with one boundary point removed, and so there is a deformation retraction from $\bar U$ onto its boundary $\partial \bar U$. Attaching $\bar U$ to $S - U$ in the obvious way to form the surface $S$, the deformation retraction $\bar U \to \partial\bar U$ induces a deformation retraction of $S$ onto $S-U$, wnich is a subcomplex of the 1-skeleton.
By the way, the subtree $T \subset S^{(1)}$ can be constructed by an explicit process. Enumerate the dual $0$-cells $v_1,v_2,\ldots \in S^{(1)}$. Construct one-ended subtrees $T_1,T_2,\ldots \subset S^{(1)}$ as follows. $T_1$ is any proper ray based at $v_1$. If $v_n \in T_{n-1}$ then $T_n = T_{n-1}$. If $v_n \not\in T_{n-1}$, let $T_n$ be the union of $T_{n-1}$ with any arc $\alpha \subset S^{(1)}$ having one endpoint at $v_n$ and intersecting $T_{n-1}$ in its opposite endpoint. Each $T_n$ is a one-ended tree by induction, and since the radius $r$ neighborhood of $v_1$ in $T_n$ stabilizes as $n \to \infty$, it follows that $T = \cup_n T_n$ is a one-ended subtree of $S^{(1)}$, and it is maximal because it contains each $v_i$.
I think this proof generalizes to any dimension, to give the theorem that Igor Belegradek refers to.
--- Edited to simplify and clarify the argument ---