Can we have a map from $(0,1)$ to $(0,1)$ such that the image of every open interval in $(0,1)$ is all of $(0,1)$ ?
Yes. My favorite, though it takes a little cleaning in the corners is this one. Let $x\in (0,1)$ and express it in base $3$. If there are an infinite number of $2$'s in the expansion, set $f(x)=x$ and ignore it. Otherwise, cut off all the leading digits through the last $2$ and read the resulting number in binary to get $f(x)$. Given any $y \in (0,1)$ expressed in binary and an interval $(a,b)$ we can find $c \in (a,b)$ expressed in ternary ending in $2$ that we can append $y$ to and stay in $(a,b)$