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Let $X$ be a locally compact space and $Y$ hausdorff

Let $f:X \to Y$ be a surjective, continuous and open function.

I need to show that given $K \subset Y$ compact there exists $L \subset X$ compact such that $f(L)=K$

What I tried : Given such $K$, consider $f^{-1}(K)$, for each $x \in f^{-1}(K)$ there is a compact neighbourhood $C_x$ of $x$, cover $f^{-1}(K)$ with those compact subsets. I'm trying to show that $f^{-1}(K)$ is a finite union of those subsets. Since finite union of compact is compact and $f$ is surjective I could just define $L=f^{-1}(K)$

I know that $f(C_x)$ is open and covers $K$, but having a finite amount covering $K$ doesn't mean a finite amount of those will cover $L$

I don't know how I can use the fact that $Y$ is hausdorff besides saying that $f^{-1}(K)$ is closed.

Any advice on how to proceed?

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    You generally can't take $L = f^{-1}(K)$. Consider $X = \mathbb{R}$ and $Y = \{0\}$. – 2017-02-13

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I don't know how I can use the fact that $Y$ is hausdorff besides saying that $f^{−1}(K)$ is closed.

That's the thing. Since $f^{-1}(K)$ is closed, the set $f^{-1}(K) \cap M$ is compact whenever $M$ is a compact subset of $X$.

I know that $f(C_x)$ is open and covers $K$, but having a finite amount covering $K$ doesn't mean a finite amount of those will cover $L$.

$f(C_x)$ need not be open, but that's not important, the $f(\operatorname{int} C_x)$ are open and cover $K$. But the important thing is that you don't need to cover $f^{-1}(K)$ with finitely many $C_x$ - that is often impossible - what you need is a compact $M \subseteq X$ such that $K \subseteq f(M)$. Then $L := M \cap f^{-1}(K)$ will do.

Now, from the fact that

$$\{ f(\operatorname{int} C_x) : x \in f^{-1}(K)\}$$

is an open cover of $K$, can you construct a compact $M\subseteq X$ with $K \subseteq f(M)$?