Conway base 13 function

Counterexample

Use base $13$ with the thirteen digits $0, 1, \ldots, 9, \mathtt{+}, \mathtt{-}, \mathtt{.}$, in this order. For $x \in \R$, take the base-$13$ expansion of the fractional part of $x$ that does not end in an infinite string of the last digit. Suppose that from some position on, this expansion has the form $$\mathtt{\pm}\, a_1 a_2 \ldots a_k \,\mathtt{.}\, b_1 b_2 b_3 \ldots$$ This tail consists of a sign symbol and finitely many digits $a_i \in \set{0, \ldots, 9}$. Then comes the point symbol, and after it only digits $b_j \in \set{0, \ldots, 9}$. Then set $f(x) = \pm\, a_1 a_2 \ldots a_k . b_1 b_2 b_3 \ldots$, read as an ordinary decimal number. Otherwise set $f(x) = 0$.

Each open interval $J$ contains all numbers with a fixed integer part whose expansion begins with a suitable finite block of digits. After that block, any tail can be appended. So for each real $y$, some $x \in J$ has a tail that encodes $y$, and $f(J) = \R$.