Can we construct a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that it has intermediate value property and discontinuous everywhere? 
Can we construct a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that it has intermediate value property and discontinuous everywhere?

I think it is probable because we can consider 
$$  y = 
\begin{cases}
\sin \left( \frac{1}{x} \right), & \text{if } x \neq 0, \\
0, & \text{if } x=0.
\end{cases}
$$
This function has intermediate value property but is discontinuous on $x=0$.
Inspired by this example, let $r_n$ denote the rational number,and define 
$$  y = 
\begin{cases}
\sum_{n=1}^{\infty} \frac{1}{2^n} \left| \sin \left( \frac{1}{x-r_n} \right) \right|, & \text{if } x \notin \mathbb{Q}, \\
0, & \mbox{if }x \in \mathbb{Q}.
\end{cases}
$$
It is easy to see this function is discontinuons if $x$ is not a rational number. But I can't verify its intermediate value property.

Sure.  The class of functions satisfying the conclusion of the Intermediate Value Theorem is actually vast and well-studied: such functions are called Darboux functions in honor of Jean Gaston Darboux, who showed that any derivative is such a function (the point being that not every derivative is continuous).
A standard example of an everywhere discontinuous Darboux function is Conway's base 13 function.
(Perhaps it is worth noting that the existence of such functions is not due to Conway: his is just a particularly nice, elementary example.  I believe such functions were already well known to Rene Baire, and indeed possibly to Darboux himself.)
