Attempt: $f: [-1,1] \rightarrow R$ is a continuous function with $x \in [-1,1] \setminus \{0\}$, and it is monotone increasing. Let $f(a)=-1$ and $f(b)=1$. Then, there exists a $c \in [-1,1]$ such that $f(a)<f(c)<f(b)$. Therefore, $f$ has the IVP.
Is this proof okay? I am not sure whether the question asks to show IVP for $x\in R$ or for $x\in R \setminus \{0\}$.
Thank you in advance!
On
The first thing to notice is that the enunciate is wrong: it should be $f(c)=y$. Anyway, what they want you to prove is that the preimage of an open set is an open set. This is the definition of continuity in topological spaces but is just question of culture, don't bother.
Ok, my approach: assuming $$0<a<b,$$ then the function $$f_l(x)=\left\lbrace\begin{array}[cc]\, f(x)&x\geq a\\f(a)&x<a\end{array}\right.$$ is continuous and satisfies IVP. The same goes for $a<b<0$ with $$f_r(x)=\left\lbrace\begin{array}[cc]\, f(b)&x\geq b\\f(b)&x<b\end{array}\right.$$
For $a<0<b$ you can find a $a_0$ in $(a,0)$ and a $b_0$ in $(0,b)$ such as $$f(a_0)=f(b_0)=0$$ (e.g. take $b_0=\pi/n$ for $n>\pi/b$) and use one of the previous functions depending on the sign of $y$.
$f(x)$ is not monotone increasing.
while you can still find intervals on which your function is monotone increasing, you really do not need monotonicity at all to prove your case.
All you need is intermediate value theorem on a small interval for which your function covers the interval [-1,1].
Good Luck.