I think $f(y)=c$ is a typo, so it should be $f(c)=y$. Am I right?
I know that this is the famous example of a discontinuous function with the IVT property,but how can I prove this?
My attempt: Let $a_1=2/3\pi$ and $b_1=2/\pi \rightarrow f(a_1)=-1$ and $f(b_1)=1$. Then, there exists a $c\in (a_1,b_1)$ such that $f(c)=0$ . $f(\frac {\frac {2}{3\pi}+\frac {2}{\pi}}{2})= f(\frac {4}{3\pi})=\sin (\frac {3\pi}{4})=\frac 1{\sqrt2}$. Then, choose $a_2 = \frac 2{3\pi}$ and $b_2= \frac {4}{3\pi}$, and continue this process until reach the precise result.
Is this the right approach?
The only case you have to worry about is when $a<0<b$. Note that $f$ takes values in $[-1,1]$ so $-1 \leq y \leq 1$. Take a look at the graph of $f$. You can easily see that in $(a,0)$ $f$ takes all values between -1 and 1 (infinitely many times, in fact!). Hence there exists $c \in (a,0) \subset (a,b)$ with $f(c)=y$. (You can also find a $c$ in $(0,b)$).