This does not have to be a very extravagant example just something that I can wrap my head around to have a concrete idea.
I was thinking that this could be satisfied by the function
$f: [0, 2\pi) \rightarrow B_1 = \{(x,y): x^2 + y^2 = 1\}$
$$
f(x) = (\cos(x),\sin(x))
$$
The point of discontinuity is obviously at $(1,0)$.
I was hoping to get a little help convincing myself that this is both an open and closed mapping. (Still getting use to the definition). So, my definition as stated follows.
Let $f:X \rightarrow Y$ be a function on the indicated spaces. Then $f$ is an $\bf{open}$ $\bf{function}$ or $\bf{open}$ $\bf{mapping}$ if for each open set $O$ in $X$, $f(O)$ is open in $Y$. The function $f$ is a $\bf{closed}$ $\bf{function}$ or $\bf{closed}$ $\bf{mapping}$ if for each closed set $C$ in $X$, $f(C)$ is closed in $Y$.
My idea was along the lines of the following.
Let $O$ be an open set in the interval $[0, 2\pi)$. Then $O = (a,b)$, $a>0, b<2\pi$. Suppose $U = (0, 2\pi) = \cup O$. Where $U$ is the collection of all open sets $O$. Then $f(U)$ is the unit circle excluding the point $(1,0)$ Therefore an open set.
Let $C$ be a closed set in the interval $[0, 2\pi)$. Let $\epsilon > 0$ be given. Then $C = [a, b]$ such that $a \geq 0$, $b \leq 2\pi- \epsilon$. Suppose $H = [0, 2\pi- \epsilon]= \cup C$. $H$ is the collection of all closed sets $C$. Then $f(H)$ is the approximately the unit circle radius 1 starting at the point $(1,0)$ ending and including a point before the point $(\cos(2\pi- \epsilon), \sin(2\pi- \epsilon))$. Therefore a closed set.
Thus we have found a discontinuous, closed and open map.
Best Answer
Let $X$ and $Y$ be topological spaces on the same underlying set (with at least two elements), where $Y$ carries the discrete topology and $x$ the indiscrete topology. Then the identity map $X\to Y$ is open, closed, not continuous.