How can I prove that a transfomation $f: \mathbb{R}^n \to \mathbb{R}^m$ is continuous if and only if the inverse image $f^{-1} (O)$ := {${x \in \mathbb{R}^n} : f(x) \in O$} $\subset \mathbb{R}^m$ of every open set $O \subset \mathbb{R}^n$ is also an open set?
I know that there exists a continuous function, for example $h:\mathbb{R}^n \to \mathbb{R}^m$ and an open set $O \subset \mathbb{R}^n$, so that $h(O) \subset \mathbb{R}^m$ isn't open. I could just take $h \equiv 0$ and $O = \mathbb{R}^n$, then $h(O)$ = {$0$} $\subset \mathbb{R}^m$ which isn't open. This implies that the statement doesn't work for images of open sets.
Further I know that a transformation $f: \mathbb{R}^n \to \mathbb{R}^m$ is continuous exactly when the inverse image $f^{-1}(A)$ of every closed set $A \subset \mathbb{R}^m$ is a closed set.
But that still doesn't help me much with how proving the statement on top..
First of all, usually the definition of a continuous map $f:X \to Y$ is that for every open $U \subset Y$ $f^{-1}(U)\subset X$ is open. So if you already know the statement for closed sets: Let $f: X \to Y$ be "continuous" (your closed set definition). Consider $U \subset Y$ open, then $Y\setminus U$ is closed. and hence $f^{-1}(Y\setminus U)= X \setminus f^{-1} (U) $ is closed, wich means $f^{-1}(U)$ is open.