An example of neither open nor closed set

$[0,1)$

It is not open because there is no $\epsilon > 0$ such that $(0-\epsilon,0+\epsilon) \subseteq [0,1)$.

It is not closed because $1$ is a limit point of the set which is not contained in it.

For a slightly more exotic example, the rationals, $\mathbb{Q}$.

They are not open because any interval about a rational point $r$, $(r-\epsilon,r+\epsilon)$, contains an irrational point.

They are not closed because every irrational point is the limit of a sequence of rational points. If $s$ is irrational, consider the sequence $\left\{ \dfrac{\lfloor10^n s\rfloor}{10^n} \right\}.$

Let $A = \{\frac{1}{n} : n \in \mathbb{N}\}$.

$A$ is not closed since $0$ is a limit point of $A$, but $0 \notin A$.

$A$ is not open since every ball around any point contains a point in $\mathbb{R} - A$.

Take $\mathbb{R}$ with the finite complement topology - that is, the open sets are exactly those with finite complement. Then $[0,1]$ is neither open nor closed. It is not open since $\mathbb{R}\setminus [0,1]=(-\infty,0) \cup (1,\infty)$ is not finite, and it is not closed since its complement, $(-\infty,0) \cup (1,\infty)$, is not open, as just demonstrated.

The interval $\left ( 0,1 \right )$ as a subset of $\mathbb{R}^{2}$, that is $\left \{ \left ( x,0 \right ) \in \mathbb{R}^{2}: x \in \left ( 0,1 \right )\right \}$ is neither open nor closed because none of its points are interior points and $\left ( 1,0 \right )$ is a limit point not in the set.