Discontinuous at infinitely many points While doing a worksheet on real analysis I came across the following problem.
$Q$. Let $f$ be a function defined on $[0,1]$ with the following property.
For every $y \in R$, either there is no $x$ in $[0,1]$ for which $f(x)=y$ or there are exactly two values of $x$ in $[0,1]$ for which $f(x)=y$.
(a) Prove that $f$ cannot be continuous on $[0,1]$.
(b) Construct a function $f$ which has the above property.
(c) Prove that any such function with this property has infinitely many discontinuous on $[0,1]$.
I really have absolutely no idea how to solve the problem. Even constructing the function is proving pretty difficult. Any help would be appreciated asap.

This example shows that the image of $f$ can be the whole of $\mathbb{R}$:
\begin{align*}
f(0) & = 0, \\
f(x) & = \begin{cases}
\frac{3}{2^n} - 8x &
   \left( \frac{1}{2^{n + 1}} \leqslant x < \frac{5}{2^{n + 3}} \right) \\
\frac{1}{2^{n - 2}} - 8x &
   \left( \frac{5}{2^{n + 3}} \leqslant x < \frac{3}{2^{n + 2}} \right) \\
8x - \frac{5}{2^n} &
   \left( \frac{3}{2^{n + 2}} \leqslant x < \frac{7}{2^{n + 3}} \right) \\
8x - \frac{3}{2^{n - 1}} &
   \left( \frac{7}{2^{n + 3}} \leqslant x < \frac{1}{2^n} \right),
\end{cases}
\\
f(x) & = \begin{cases}
\frac{1}{8x - 5} &
   \left( \frac{1}{2} \leqslant x < \frac{5}{8} \right) \\
\frac{1}{8x - 6} &
   \left( \frac{5}{8} \leqslant x < \frac{3}{4} \right) \\
\frac{1}{7 - 8x} &
   \left( \frac{3}{4} \leqslant x < \frac{7}{8} \right) \\
\frac{1}{8 - 8x} &
   \left( \frac{7}{8} \leqslant x < 1 \right),
\end{cases} \\
f(1) & = 0.
\end{align*}
(The variable $n$ takes all positive integral values.)
\begin{alignat*}{2}
f\left(\left[\frac{1}{2^{n + 1}}, \frac{5}{2^{n + 3}}\right)\right) & =
f\left(\left[\frac{5}{2^{n + 3}}, \frac{3}{2^{n + 2}}\right)\right) && =
\left(-\frac{1}{2^{n - 1}}, -\frac{1}{2^n}\right], \\
f\left(\left[\frac{3}{2^{n + 2}}, \frac{7}{2^{n + 3}}\right)\right) & =
f\left(\left[\frac{7}{2^{n + 3}}, \frac{1}{2^n}\right)\right) && =
\left[\frac{1}{2^n}, \frac{1}{2^{n - 1}}\right),
\end{alignat*}
\begin{alignat*}{2}
f\left(\left[\frac{1}{2}, \frac{5}{8}\right)\right) & =
f\left(\left[\frac{5}{8}, \frac{3}{4}\right)\right) && =
\left(-\infty, -1\right], \\ \\
f\left(\left[\frac{3}{4}, \frac{7}{8}\right)\right) & =
f\left(\left[\frac{7}{8}, 1\right)\right) && =
\left[1, \infty\right).
\end{alignat*}
Amusingly, $f$ is continuous at $0$.
