How to prove the topological space is not metrizable？

If this topology is coming from a metric on $Z$, then it has to be Hausdorff. But you can easily show that this topology is not Hausdorff. For example take two points 1 and 2 in $Z$. every open set containing $1$ contains 2 as well.

A metrizable topological space is in particular a $T_1$-space, i.e. for all $n,m \in \mathbb{N}$, $n \not=m$, there have to exist open sets $U,V \in \lambda$ such that $n \in U$, $m \in V$, $U \cap V = \emptyset$.

But this is clearly not the case: Consider $n := 1$ and $m \in \mathbb{N}$, $m>1$ arbritrary. The set $O_1 = \{n \in \mathbb{N}; n \geq 1\}=\mathbb{N}$ is the only open set containing $1$, so $U=O_1=\mathbb{N}$. But since this implies $m \in U$, we have $$m \in U \cap V$$ i.e $U \cap V \not= \emptyset$ for all $V \in \lambda$ such that $m \in V$.

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Concerning your own proof: The idea is correct. Assume that the topology is generated by some metric $d$. Then $B(1,r)$ is a open set for all $r>0$, so there has to exist in particular a set $O \in \lambda$ such that $O=B(1,\varepsilon)$ where $\varepsilon:=d(1,2)$. By definition, we have $2 \notin O$. But since $O=O_1 = \mathbb{N}$ is the only set in $\lambda$ containing $1$, this is clearly a contradiction.