I asked myself:
If $f: \mathbb R \to \mathbb R$ and $f(\mathbb R)$ is connected then does $f$ map connected sets to connected sets?
My idea is that is true: There exists $f: \mathbb R \to \mathbb R$ with $f(\mathbb R)$ connected and a connected set $A \subseteq \mathbb R$ with $f(A)$ not conneted.
I try to find an example of this but failed. Now I start to think maybe it is wrong. But I don't know how to think about it. Please can somebody help me?
On
It is not true in the general case, but it's true for continuous functions (maybe that's why you didn't find a counterexample, sometimes all the functions you think of are continous).
For example, consider $f(x) = \cases{x & $x\leq 0$\\ 1-x & $0<x\leq 1$ \\ x-1 & $1<x$}$
This function is surjective (i.e. $f(\mathbb R) = \mathbb R$), but the image of $(-\frac{1}{2}, \frac{1}{2})$ is not connected, since it is $(-\frac{1}{2}, 0] \cup (\frac{1}{2}, 1)$.
Let $f(x) = x$ for $x \ge 0$ and $f(x) = x+1$ for $x < 0$. Then $f(\mathbb R)$ is connected, but $f([-1/10, 1/10])$ is not.