If I understood the OP correctly, he wants some simple examples of functions, which are not continuous and they have Darboux property. (He wants to practice showing that a function has intermediate value property on some concrete examples.)
I've given a few examples. I have made this post CW, so feel free to add further examples.
Functions which are not continuous, but are derivatives:
$f(x)=
\begin{cases}
\sin\frac1x, & x\ne 0, \\
0 & \text{otherwise}.
\end{cases}
$
$g(x)=
\begin{cases}
2x\cos\frac1x+\sin\frac1x, & x\ne 0, \\
0 & \text{otherwise}.
\end{cases}
$
$h(x)=
\begin{cases}
2x\sin\frac1{x^2}-2\frac1x\cos\frac1{x^2}, & x\ne 0, \\
0 & \text{otherwise}.
\end{cases}
$
The functions $f(x)$, $g(x)$ are $h(x)$ are from the book Van Rooij-Schikhof: A Second Course in Real Analysis (in the Introduction.).
Functions which are not continuous, but have Darboux property (intermediate value property):
$f_2(x)=
\begin{cases}
\sin\frac1x, & x\ne 0, \\
1 & \text{otherwise}.
\end{cases}
$
Again from the book Van Rooij-Schikhof: A Second Course in Real Analysis (in the Introduction.).
Consider the function $g\colon [-1,1] \to \mathbb R$ defined by $g(x) = f(x) - x$. Notice that since $f(x)$ and $x$ are continuous on $[-1,1]$, we know that $g$ is also continuous on $[-1,1]$. Furthermore, observe that:
\begin{align*}
g(-1) &= f(-1) - (-1) > -1 + 1 = 0 \\
g(1) &= f(1) - 1 < 1 - 1 = 0
\end{align*}
Hence, since $g(1) < 0 < g(-1)$, it follows by the Intermediate Value Theorem that there exists some $c \in (-1,1)$ such that $g(c) = 0 \iff f(c) - c = 0 \iff f(c) = c$. So $c$ is a fixed point of $f$, as desired.
Best Answer
Hint: Look for a function with a jump discontinuity.