Find a function defined on $[0,1]$, valued on an interval, but is discontinuous at each point.
That is, try to find a function $f: [0,1]\to \Bbb R$ such that $f([0,1])$ is an interval, but discontinuous at each point.
Here is my try. Let $\Bbb Q=\{r_1,r_2,\cdots\}$ with $r_1=0, r_2=1, \cdots$. Then define $f(x)=x$ for irrational $x$, $f(r_n)=r_{n+1}$, then $f([0,1])=(0,1]$, and is discontinuous at rational points, but how about irrational points?
Try $$ f(x) = \cases {x & if $x$ is rational\cr x + 1/2 & if $x < 1/2$ and $x$ is irrational\cr x - 1/2 & if $x > 1/2$ and $x$ is irrational}$$