What we need from the map is that it preserves both compactness and openness. The continuity guarantees that if $x$ is contained in a compact subspace then its image is contained in a compact subspace as well. The openness of the map guarantees that a neighborhood of $x$ within the compact subspace maps into a neighborhood of the image within the compact subspace containing it. If a continuous function is not open, in general, we cannot guarantee the second property. To find a counterexample we need to consider some space which is locally compact but not compact (otherwise the continuity will be enough). Moreover, it must be mapped onto a space that is not locally compact. By now I know three types of spaces which are not locally compact: $\mathbb{R}^\omega$ (infinite products), $[0,1]\cap\mathbb{Q}$ (infinite totally disconnected) and $[0,1]\subset\mathbb{R}_l$ (strictly finer than a compact Hausdorff topology). I think the second one should work. So we should take a locally compact but not compact space, for example, a locally compact unbounded subset of $\mathbb{R}$ and construct a continuous function that would map it onto $[0,1]\cap\mathbb{Q}$ . Let $A=\cup_n(n,n+1)$ and $[0,1]\cap\mathbb{Q}=\{q_1,q_2, . . .\}$ . Let $f((n,n+1))=q_n$ . Then $f$ is a continuous (the preimage of any set is the union of open intervals) non-open (the image of a bounded open interval is not open in the range) map defined on a locally compact set $A$ whose image is exactly $[0,1]\cap\mathbb{Q}$ which is not locally compact. Cool!