Firstly, the task:
Let $X$ be locally compact space, $Y$ be a Hausdorff space, $f:X \rightarrow Y$ be a continuous open surjection and let $K$ be a compact set in space $Y$. Prove that there exists a compact set $C$ in space $X$ such that $f(C)=K$.
I think that I have a proof, but it seems to use the axiom of choice. I have tried, but I can't decide whether I really use that axiom or not. I need a proof that doesn't use it. Please, help me prove it without that axiom, or explain me that I haven't really used it.
Sketch of my (incorrect) proof:
Since $f$ is surjective, for each $y \in Y$ there is some $x_{y} \in X$ such that $f(x_y)=y$. Let's choose $C=\{x_y | y \in Y\}$. Now $f$ maps each open covering of $C$ to an open covering of $K$ and that one has a finite subcovering. Now for every $y \in K$ there exists some open set in that subcovering that contains it, so the corresponding open set in the initial covering of $C$ contained $x_y$. The collection of all the corresponding open sets (the sets whose images are in the subcover of $K$) is finite and cover whole $C$, since elements $x_y$ are unique among those that map to a chosen $y \in Y$, so we can conclude that $C$ is compact too.
I doubt that this is correct, since I used only one of four conditions, and I think that it is because I use the axiom of choice.
Best Answer
Your proof isn't correct, indeed. The subtle mistake is that, since $f$ is not (in general) injective, if $y\in f(U)$ you can't argue that the $x_y \in U$.
HINT: Every point of $f^{-1}(K)$ admits a compact neighborhood. Use these compact neighborhoods to get an open covering of $K$. (Keep in mind that these neighbourhoods are subsets of $X$ whereas $K$ is a subset of $Y\dots$).