Solutions

a) The function $f(x)$ is not continuous on $[-1,1]$ because its limit at $x=0$ does not exist. However, $f$ is continuous on $[-1,0)$ and $(0,1]$, so we can apply the IVT on these intervals. For intervals $[a,b]$ with $a\le0\le b$, $f$ takes on all values in $[-1,1]$, so it surely takes all values between $f(a)$ and $f(b)$.

b) For contradiction, suppose $f$ is not continuous. Consider the point $x=a$. This means that there exists $\varepsilon>0$ such that for all $\delta>0$, there is an $x$ with $|x-a|<\delta$ and $|f(x)-f(a)|\ge\varepsilon$. Without loss of generality, suppose

$$f(x)-f(a)\ge\varepsilon,$$

so that $f(x)\ge f(a)+\varepsilon$, and $x>a$. By the IVT, there exists $\beta$ such that $f(\beta)=c$ for $f(a)<c<f(a)+\varepsilon$. Now, we can find another point $y$ by shrinking $\delta$ so as to exclude the points $x$ and $\beta$ from $[y-\delta,y+\delta]$. In this interval, we also have $|f(y)-f(a)|\ge\varepsilon$. But if $f(y)-f(a)\ge\varepsilon$, then $f(y)\ge f(a)+\varepsilon$, and by the IVT, $f$ takes on the same values in $(y,\beta)$ as it does in $(\beta,x)$. A similar contradiction arises if $f(a)-f(y)\ge\varepsilon$ instead.

c) We can repeat the interval-generating process in b), where $f$ repeatedly takes on the same values, thus contradicting the hypothesis.