Can we find a function $f:\mathbb{R}\rightarrow \mathbb{R}$ that is open, closed, but not continuous?

No such function exists.

Assume that $f((a,b))$ is a bounded open set. Then $f((a,b))= (c,d)$ for some $c,d\in \mathbb{R}$. Indeed, $f((a,b))$ is an open set hence of the form $\bigcup (a_n,b_n)$ however $f([a,b])$ has at most two more points than $f((a,b))$, so we conclude.

Claim: if $f((a,b))= (c,d)$ then $f$ is continuous and strictly monotonic on $(a,b)$.

Proof: Firstly, we will show that $f$ is injective. Indeed assume that $f(x_1)= f(x_2)$ for some $x_1,x_2\in (a,b)$. Then take the $f((x_1,x_2))=(c_1,d_1) $ for some $c_1,d_1$, a contradiction since $f( [x_1,x_2] )$ needs two points to be closed. Now, since $f$ has the IVT propoerty and it is injective it is strictly monotonic and continuous.

We will argue by contradiction. Assume that $f$ is not continuous at $x_0$, and set $V_n=( x_0-1/n,x_0+1/n )$. Now, the claim implies there are $z_n<w_n$ such that $f(V_n )=(-\infty ,z_n)\bigcup (w_n,\infty) $ where $z_n$ or $w_n$ can possibly take the value $\pm \infty$. Also, notice that $w_n$ is increasing and $z_n$ is decreasing.

First case, one of the $z_n,w_n$ converges (to a finite limit). WLOG assume that $w_n \rightarrow w$. Now, pick a number $\beta$ in $(w,\infty)$ that is different from $f(x_0)$. We can find a sequence $p_n\in V_{n-1}\setminus V_n$ such that $f(p_n)=\beta +1/n$. A contradiction since the closed set $\{x_0\}\bigcup \{p_n|n\in \mathbb{N}\}$ gets mapped via $f$ to $\{f(x_0)\}\bigcup \{\beta +1/n|n\in \mathbb{N}\}$ which is not closed.

In the second case, we have $z_n \rightarrow -\infty$ and $w_n \rightarrow\infty$. This is a contradiction since the previous implies $\bigcap_{n \in \mathbb{N}}f(V_n)= \emptyset$, however $\bigcap_{n \in \mathbb{N}}f(V_n)$ always contains $f(x_0)$.