In most applications, topological spaces are assumed to be Hausdorff ($T_2$), i.e. that any two distinct points have disjoint neighbourhoods. This is true among others for metric spaces, like the real numbers, and it immediately implies the $T_1$ condition which says that for any pair of distinct points, each has a neighbourhood not containing the other, which is actually equivalent to saying that any singleton set is closed.
Now, finite unions of closed sets are closed, so if a topological space $X$ is $T_1$, then any finite set is closed. On the other hand, if a space is not $T_1$, then if you take the pair of points witnessing that, i.e. $x,y\in X$ such that for any open $U\ni y$ we have $x\in U$, then the set $\{x\}$ will not be closed, an in fact any set containing $x$ but not containing $y$, finite or not, will not be closed. In particular, any finite set which is not open and does not contain $y$ is neither open nor closed.
For example, any nonempty finite subset of an infinite space with trivial topology is neither open nor closed.
Note that even if a space is not $T_1$, it can still be the case that every finite set is either open or closed, for example if you take $X=\{x,y\}$ with open sets $\emptyset,X,\{x\}$, then $\{y\}$ is closed and all other subsets of $X$ are open.
Consider $B_r(1)$. You should notice for any $r>0$, $B_r(1)$ contains both point in $(0,1]$ and not in $(0,1]$. This implies that neither $(0,1]$ or its complement are open.
Best Answer
$[0,1)$
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.