Suppose that $f: \mathbb R^n \to \mathbb R^n$ is a bijection and $n\geq2$. Can $f$ send every open set onto non-open set? 
Suppose that $f: \mathbb R^n \to \mathbb R^n$ is a bijection and $n\geq2$. Can $f$ send every open set onto non-open set?

I do not know what exactly to write about this problem. Did I try anything? No, I do not have a good idea. Why am I asking this? Because here is this very highly upvoted question of Willie Wong that I think of, and, it seems to me as a good start to investigate what exactly can bijections "do" and what they can´t, so, as a starting point, I decided to ask this question.
Edit: Thomas wrote a useful comment that $\emptyset$ and $\mathbb R^n$ are mapped onto open sets. So to exclude trivialities, suppose that from consideration we exclude the empty set and the whole space $\mathbb R^n$, to make this more interesting.
 A: The set $\mathbb R^n\setminus\{\mathbf v\}$ is always open, and always goes to the open set, $\mathbb R^n\setminus\{f(\mathbf v)\}.$
More generally, if $F\subseteq \mathbb R^n$ is finite then $\mathbb R^n\setminus F$ is open, and its image $\mathbb R^n\setminus f(F)$ is open. 
A: Reading over the OP's question and their comments, it becomes evident that they are interested in 'scramble-it-up' bijections and @AlexanderGeldhof's comment. Here we want to offer some food for thought to the OP's curiosity; we'll only be examining bijections $f: \Bbb R \to \Bbb R$.
The first thing to note, is that if $\tau: \Bbb R \to G$ is a bijection onto a set $G$, then every bijection $g: G \to G$ can be mapped to a bijection
$${\tau}^{-1} \circ g \circ \tau: \Bbb R \to \Bbb R$$
and every bijection mapping $\Bbb R$ to $\Bbb R$ has this form.
A second point of interest is that by using the axiom of choice, the existence of bijections can be postulated yet you can't specify an algorithim to 'pin things down'; see
Hamel Basis
You can 'accept' the existence of bijective linear transformations of $\Bbb R$ over $\Bbb Q$ by matching up any two different Hamel Bases, but don't 'strain your brain' trying to 'see them' (c.f. this).
Finally, it seems only fair to define a transformation on $\Bbb R$ that 'rips apart' bounded open intervals.
Define $f\colon\mathbb{R}\to\mathbb{R}$ as follows:
$f(x) = \left\{\begin{array}{lr}
        \;\;\;x+1\, \;\;\;\text{ |} & \text{when } x \text{ is a rational number}\\
        \;\;\;x-1\, \;\;\;\text{ |} & \text{when } x \text{ is an irrational number}
        \end{array}\right\}$
