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.
Concerning the usage of "if":
In definitions people usually write X is called Y if Z holds, even though they mean "if and only if". Since it is a definition, there is no other object with that name and other probably weaker definitions.
Concerning "if", "iff", and $\implies$:
"If condition A holds, then statement B is valid" is usually written as $A \implies B$ and read as $A$ implies $B$. "Iff", short for "if and only if", means $A \implies B$ and $B\implies A$ are true at the same time.
Concerning the characterizations of continuity:
Here, $f^{-1}$ means the set of all inverse images:
$$ f^{-1}(V) = \{ x\in X \mid f(x) \in V \}. $$
It is always defined, but might be empty.
Definitions 1 and 2 are the same in metric space. To see this, each neighborhood of $a$ contains some $\delta$-neighborhood of $a$ as subset.
Definition 3 states that $f$ is continuous at any $a\in X$, not only at a particular one, like in definition 2. To see that "continuous at every $a\in X$" as in definition 2 is the same as definition 3, a set $S\subseteq X$ is open if and only if each $x\in S$ is an interior point of $S$.
Concerning your proof:
For definition 1, you're are absolutely right. In a formal proof however, for any given $\varepsilon > 0$ you have to pick a $\delta > 0$ (may depending on $\varepsilon$).
For definition 2, you have a typo there. $\{ \| x \| \mid x\in B(a,\delta) \}$ is a subset of $(\|a\| - \varepsilon, \|a\| + \varepsilon)$.
Best Answer
In a metric space we begin with the $\epsilon$-$\delta$ definition of continuity, which is then globalized by the requirement that the $\epsilon$-$\delta$ holds at every point. This is a natural starting point because from the quantitative $\epsilon$-$\delta$ relation we are led to several related important concepts:
None of the above related concepts exist in a topological space, where the notion of continuity is not quantifiable.
To define continuity in a topological space we operate not with a pair of nearby points $x_1,x_2$, but with a set of points; thus, the global picture emerges at once. "Preimage of an open set is open", a standard way to define continuity in a topological space, is naturally global. Equivalently, one can state it as "preimage of a closed set is closed". From here, your definition of continuity at a point, $$A \in \mathcal{P}(X), \, x_0\in \overline{A}\implies f(x_0)\in \overline{f(A)}\tag{1}$$ is obtained by localizing a global definition : you have localized the property of "containing all its limit points" by focusing on a particular limit point. As a result, (1) is more contrived than the global definition. If you try to reprove the basic results of point-set topology always using (1) as definition of continuity, you will likely find that the proofs become more cumbersome.
I prefer to read "not such a useful" as "not such a natural"; which however amounts to the same thing.