$B(a,r)$ is just the solution of $\left|\dfrac1a -\dfrac1x\right|<r$, which is easily solved:
$
x>\dfrac{a}{1+a r} \qquad \qquad \qquad\text{if } r\ge\dfrac1a
$
$
\dfrac{a}{1+a r}<x<\dfrac{a}{1-a r} \qquad \text{if } r<\dfrac1a
$
In the first case, the ball is infinite. In the second case, the ball is finite.
In particular,
the ball is infinite if $r\ge1$ and the ball centered at $1$ is all of $\mathbb N$ if $r=1$.
Another approach:
Suppose we have some metric space $(M,d)$, and a subset $M' \subset M$. Then $(M',d)$ is a metric space as well. Then a set $U' \subset M'$ is open iff there is some open $U \subset M$ such that $U'=U \cap M'$. This is fairly straightforward to prove. It is also straightforward to prove the corresponding result for closed sets.
In your examples, $M=\mathbb{R}$ with the usual metric and $M'=(-1,1]$. So, your examples can be written as:
(i) $(-1,1] = \mathbb{R} \cap M'$, so $(-1,1]$ is both open and closed in $Y$.
(ii) Needs a little more attention. If $(-\frac{1}{2}, 0] = U \cap M'$, where $U$ is open in $\mathbb{R}$, then we would also have $(-\epsilon,\epsilon) \subset (-\frac{1}{2}, 0]$ for some $\epsilon>0$, so it cannot be open. Similarly, if $(-\frac{1}{2}, 0] = C \cap M'$, where $C$ is closed in $\mathbb{R}$, then we would also have $\frac{1}{2} \in (-\frac{1}{2}, 0]$, so it cannot be closed.
(iii) $(-1,0] = (-\infty,0] \cap M'$, so $(-1,0]$ is closed in $M'$, and the same line of reasoning for (ii) shows that it cannot be open in $M'$.
Best Answer
In a metric space, every open ball is an open set, but certainly not the other way around. A trivial example is simply to take the union of two disjoint open balls (in, say $\mathbb R^2$ if you want to get a nice picture). Most certainly the union of two disjoint open balls is not an open ball, but it is an open set.
To solve your problem, think of closure properties of $\sigma$-algebras, of open sets and open balls (in separated metric spaces), and the relationship between open/close and set complementation.