A key property that all metric spaces have is the Hausdorff axiom. Namely given any $x,y \in X$ such that $x \neq y$ there are open sets $U$ about $x$, $V$ about $y$ such that $U \cap V = \emptyset$.
To see how this holds in metric space take any two distinct points $v$ and $w$. They are distinct so by definition of a metric $d(v,w) = \epsilon$ for some $\epsilon > 0$. By the Archimedean property of the reals there exists $n \in \Bbb{N}$ such that
$$ 0 < \frac{1}{n} < \epsilon$$
from which it follows that $B_{\frac{1}{n}}(v)$ does not intersect $B_{\epsilon - \frac{1}{n}} (w)$. Since $v,w$ were any two distinct points in your metric space this proves the claim.
So to show that your topological space above cannot be turned into a metric space it is sufficient to show that there are positive integers $x,y \in Z$ such that any two open neighbourhoods about $x$ and $y$ respectively must have non-trivial intersection. Consider $1$ and $2$ that are in $\Bbb{Z}$. Then it is easy to see that the only open neighbourhood about $1$ is $O_1 = \{1,2,3, \ldots \}$ while the only open neighbourhood about $2$ is $O_2 = \{2,3,\dots \}$ or even $O_1$ as well. However it is clear that
$$O_1 \cap O_1 \neq \emptyset, \hspace{5mm} O_1 \cap O_2 \neq \emptyset$$
from which it follows that the topology $\mathcal{J}$ that you put on $Z$ is not Hausdorff and hence is not metrisable.
There exist a metric space $X$ containing two open balls $B_1$ and $B_2$ of radii $r_1$ and $r_2$ respectively such that $B_1 \subsetneq B_2$ and $r_1>r_2$.
For example, take $X=[0,+ \infty)$ and $B_1=B(0,1)=[0,1)$ and $B_2=B(1/3,4/5)=[0,1+2/15)$.
Best Answer
1) Co- countable topology 2) Co- finite topology 3) Sierpiński spaces are example of non metrizable topological spaces.
Normally topology just comes from the basis[generator of a topology], but metric spaces come from the distance function $d$. Then we observe that the open balls gives us the basis & the topology generated from that is the metrizable topology.
Again see all the topological properties concerning open sets such as arbit union of open sets are open, convergence of sequence, continuity[with taking distance in the role] etc hold for both spaces metrizable & non metrizable topological spaces , where the difference lies is concerning distance such as Hausdorff property, $T_1,T_3 $ properties etc.