$[0,1]\times\mathbb{N}/(0,k)$ not metrizable. $X = [0,1]\times\mathbb{N}/((0,1)\sim(0,2)\sim\dots)$.
I read that $X$ is not metrizable since sequence $\{(\frac{1}{n},n)\}$ is closed in $X$ and therefore does'n have limit. 
But i don't understand why it's don't converges to zero. 
For any $\varepsilon > 0$ there $N$ such $n > N \implies x_n \in [0,\varepsilon)\times \mathbb{N}$.
 A: The problem here is that the zero has much smaller neighborhoods than $[0,\varepsilon)\times\mathbb{N}$. This is namely because we are looking at a product space of two sets and $\mathbb{N}$ has the discrete topology, i.e. every singleton in $\mathbb{N}$ is open. Even if the sequence is eventually in the neighborhood $[0,\varepsilon)\times\mathbb{N}$ for any $\varepsilon>0$, it doesn't necessarily guarantee convergence.
Lets denote $Y=[0,1]\times \mathbb{N}$. The basis for the product topology are sets of the form $I\times \{n\}$, where $I\subseteq [0,1]$ is an open interval and $n\in\mathbb{N}$. Note in particular that $[0,\frac{1}{2n})\times\{n\}$ is an open subset of $Y$ for any $n\in\mathbb{N}$. Hence 
\begin{align*}
U:=\bigcup_{n\in\mathbb{N}}\Big([0,\frac{1}{2n})\times\{n\}\Big)
\end{align*}
is an open subset of $Y$, and $x_n\notin U$ for all $n\in\mathbb{N}$.
Now if $q:Y\to X$ denotes the quotient map, a subset $V$ of the quotient space $X$ is open if and only if $q^{-1}(V)\subseteq Y$ is open. Now note that $q(U)$ is an open subset of $X$ because $(0,n)\in U$ for all $n\in\mathbb{N}$ so $q^{-1}(q(U))=U$. In particular, $q(U)$ is an open neighborhood of the origin, i.e. the equivalence class of the point $(0,1)$. Since $x_n\notin q(U)$ for all $n\in\mathbb{N}$ then $(x_n)$ does not converge to the origin.
