Compact image is compact? 
Let $f: X \to Y$ be a function between metric spaces $X, \phi$ and $Y, d$ such that $f^{-1}(U)$ is open in $X$ for every subset $U$ open in $Y$. Prove that if $C$ is a compact subset of $X$, then $f(C)$ is a compact subset of $Y$.

My attempt: Let $C$ be a compact subset in $X, \phi$. Let $\{V_{\alpha}\}$ be a collection of open sets such that $C^{cpct} \subset \bigcup_{\alpha} V_{\alpha}$. Then there exist $\alpha = 1, 2, ..., n$ such that $C^{cpct} \subset \bigcup_{\alpha} V_{\alpha_n} : n \in \mathbb{N}$. 
I want to show this subcover is open and thus it has an open image, but I'm not sure as if that's correct because only the converse is true. Is it true?
I need way better notation. Also, how to continue - any suggestions?

The assumption that $f^{-1}(U)$ is open in $X$ if $U$ is open in $Y$ tells us that the function is continuous. But it doesn't tell us that the images of open sets are open (i.e., $f$ is not necessarily an open map). But all you need to know for the proof is that inverse images of open sets are open.
Try to prove the contrapositive statement:
Assume $f(C$) is not compact. So there exists an irreducible open cover $\displaystyle\bigcup_\alpha U_\alpha$ of $f(C)$. Now consider $\displaystyle\bigcup_\alpha f^{-1}(U_\alpha)$. With a little work, you'll see that this is an irreducible cover of $C$, and so $C$ is not compact.
