I would like to know if there exists a differentiable vector-valued function $f:[a,b]\rightarrow \Bbb R^2$ such that the image $f'([a,b])$ of its derivative is disconnected.
My Attempt
First of all I asked for $f$ to be vector-valued because, if it were real-valued, then Darboux's Theorem would imply that $f'([a,b])$ be connected. For a similar reason, if we let $x(t)$ and $y(t)$ denote the component functions of $f$, then their derivatives can not both be continuous, for otherwise $f'$ would also be continuous hence preserving connectedness. Therefore, WLOG, $x'$ has to be discontinuous; however, Darboux's Theorem imply that it can not have discontinuities of the I kind or discontinuities of the II kind with infinite directional limits on $]a, b[$. Now, all the functions I came up for $x$ are not enough to construct the desired $f$ and I am starting to believe that such a function doesn't actually exist but I am not able to prove it.
Any help or hint, as always, is highly appreciated!