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?
 A: 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.
A: 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.
A: We can see that map $f$ is a continuous , for a continuous i have some equivalent conditions in metric space: 
$a)$ $f$ is a continuous map 
$b)$ The image of an open set is an open set
$c)$ The image of an close set is an close set .
So $f$ is continuous , i will show that if $C$ is a compact set then $f(C)$ is a compact set , I choose an any infinity sequence $f(x_1),f(x_2),..f(x_n),,,$ in $f(C)$ , because $C$ is compact so i can choose $(x_{k_n})$ of $(x_n)$ that converging and $f$ is continuous then $f(x_{k_n})$ is converging . Finally i have $f(C)$ is compact set 
