I've been reading about Darboux functions recently and every example of a everywhere discontinuous Darboux function is also a function such that the image of every non-empty open interval is the image of the entire function. The canonical example being the Conway base-13 function. We will call these functions "constant image" functions for convenience. I was wondering if there were any Darboux functions that are everywhere discontinuous but are not constant image functions.
After discussing this with a colleague we came up with a function that seems to be a Darboux function that does not have this property. Here is a definition of the function we came up with:
Let us define a function $f$. This function will be similar to the Conway base-13 function. We first express the input in base-11 with digits $0-9,A$ such that it does not end in an infinite number of trailing $A$s. If the expansion ends with $A y_1 y_2 y_3 y_4\dots$ where $\forall n:y_n \neq A$ then the function will output $0.y_1 y_2 y_3 y_4\dots$ as a base-10 number.
This function $f$ is a such that the image of every non-empty open interval is the image of the entire function with an image of $[0,1]$, we now make a new function $$ g(x)=f(x)+x $$
This discontinuous everywhere function clearly cannot has an unbounded image but every bounded interval on this function has a bounded image. So it cannot satisfy our property but it appears, intuitively, to be a Darboux function. However a proof that this function is Darboux has escaped us.