How to prove the topological space is not metrizable？ Let $Z$ be the set of positive integers. For each positive integer n, let $O_{n}=\{ n,n+1,n+2,\dots \}$.Let $λ=\{\emptyset,O_{1},O_{2}, \dots, O_{n}.\dots\}$. Then $(Z,λ)$ is a topological space but not metrizable.
My proof is: Let $d(1,2)=\varepsilon_{1，2}$ , if $\forall x \in Z$, $d(1,2) \leq d(1,x)$ then choose $\delta < \varepsilon_{1，2}$, $B(1,\delta)$ only contains 1, and there does not exist a subset $\{1\}$ in λ,
if $\exists x\in Z$ ，st $d(1,x)<d(1,2)$ then $B(1,\delta)$ does not contain 2 and 1 is in the set, and there is not a such set.
so $(Z,λ)$ is not metrizable.
Is this proof correct？Is the metrizable topological space arise from the metric space  have the same open set？ I read a book about topology but it does not explain it and I got confused
 A: 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: 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 $m \in V$.

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.
