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.).
The function
$$f(x) = \begin{cases}10 & x = -1\\ 0 & x\in (-1,1) \\ 20 & x=1\end{cases}$$ does not satisfy the IVT. There is no $x\in(-1,1)$ such that $f(x)$ is between $f(-1)$ and $f(1)$.
However, clearly $f$ is continuous on $(-1,1)$.
Best Answer
A monotonic function can have only discontinuities of first kind or simple discontinuity.
The idea of the proof is that if $p$ is a point of discontinuity, then $f(p-)$ and $f(p+)$ exists and $f(p-)<f(p+)$. We can easily show that $$\color{red}l=\sup\limits_{a<t<p}f(t)=f(p-)<f(p+)=\inf\limits_{p<t<b}f(t)=\color{red}L.$$
Now using definition of $\sup$ and $\inf$, choose some $\varepsilon>0$ and there exists $r$ and $s$ such that $$ r<x_1<p\Rightarrow l-\varepsilon<f(x_1)\le l\\ p<x_2<s \Rightarrow L\le f(x_2)<L+\varepsilon$$
Observe that $f(x_1)<\dfrac{l+L}{2}<f(x_2)$ then by intermediate value property, there should be some $c\in (x_1,x_2)$ such that $f(c)=\dfrac{l+L}{2}$. A contradiction!