real analysis - Discontinuous everywhere but range is an interval

Does there exist a function which is discontinuous everywhere but range set is an interval.

Answer

Yes. We have $f:[0,1]\to [0,1]$ given by

$$f(x) =\cases{x& if $x$ is rational \\ x+\frac12& if $x\lt \frac12$ is irrational \\ x-\frac12 & if $x\gt\frac12$ is irrational}$$ For any $x\in[0,1]$ it's easy to see that $f(x)\in[0,1]$ as well. For any $y\in [0,1]$, we have $f(y) =y$ if $y$ is rational, otherwise $f(y\pm 1/2) =y$, with sign depending on whether $y\gt1/2$ or not. So the unit interval is the range set.

As for discontinuity, let $0<\epsilon <1/2$. Then there is no $\delta$ small enough, no matter what $x$ is, since in any open interval containing $x$ there are both rational and irrational numbers.