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?

You wish to show the image of a compact set is compact under a continuous mapping.  This can be done by taking finite sub-covers.  Let $\{\mathcal{O}_\alpha\}_{\alpha\in A}$ be an open cover of $f(C)$. Then $\{\mathcal T_\alpha\}_{\alpha \in A}$ is an open covering of $C$, where $T_\alpha \equiv f^{-1}(\mathcal{O}_\alpha)$ for each $\alpha \in A$.
Now choose $\alpha_1 \ldots \alpha_n$ for some $n \in \mathbb{N}$ such that $C \subset \cup_{i=1}^n T_{\alpha_i}$.  This can be done since $C$ is compact. 
I will leave it to you to show that $\cup_1^n f(T_{\alpha_i}) = \cup_1^n \mathcal{O}_{\alpha_i} \supset f(C)$.  
Since we have taken an arbitrary open cover of $f(C)$ and found a finite subcover, then $f(C)$ must be compact.
Since we have chosen an arbitrary compact set $C$ and shown $f(C)$ is compact, then the image of any compact set is compact under $f$. 
And finally, since $f$ is an arbitrary continuous function, we have shown that continuous functions map compact sets to compact sets.
