No series converges towards $(0,0) $ in $(\mathbb{N}\times\mathbb{N})\setminus (0,0)$ with a non standard topology in my topology course we are going through sequences and convergence of sequence and one exercise in my book says the following. 
Let $X=(\mathbb{N}\times\mathbb{N})\cup \{(0,0)\}$ with the following topology: (i) $X\setminus\{(0,0)\}$ has the discrete topology; (ii) $U$ is a neighbourhood of $(0,0)$ if $(0,0)\in U$ and the set $\{n\in\mathbb{N}\cup\{0\} \mid (n,m)\not\in U\}$ is finite for almost all $m\in\mathbb{N}\cup\{0\}$. Show that 
(a) $X$ is Hausdorff,
(b) there is no sequence in $X\setminus\{(0,0)\}$ convergent to $(0,0)$. 
Part (a) is clear to me, but part (b) is a bit obscure. I understand it has to do with the strange way the neighborhood of $(0,0)$ and I tried to argue by contradiction but didn't reach my goal. Can anybody help me? 
 A: Let $S=(p_x)_{x\in \mathbb N}$ be a sequence in $\mathbb N^2.$ Let $T=\{p_x: x\in \mathbb N\}.$  
If $T\cap (\mathbb N \times \{m\})$ is finite for every $m\in \mathbb N$ then $\{(0,0)\}\cup (\mathbb N^2$ \ $T)$ is a nbhd of $(0,0)$ that contains no terms of $S$. 
If  $T\cap (\mathbb N \times \{m_0\})$ is infinite for some $m_0\in \mathbb N,$ then  there are infinitely many $x\in \mathbb N$ such that $p_x\in \mathbb N \times \{m_0\},$ and so $U=\{0,0\}\cup (\mathbb N^2$ \ $(\mathbb N\times \{m_0\})$ is a nbhd of  $(0,0)$ such that $p_x\not \in U$ for infinitely many $x\in \mathbb N.$
BTW this shows that this topology is not metrizable. Because $\{(0,0)\}$ is not open, but if $p\in Y$ where $Y$ is metrizable and $\{p\}$ is not open, then there exists  a sequence $(p_x)_{x\in \mathbb N}$ in $Y$ \  $\{p\}$ such that $\{x\in \mathbb N:p_x\not \in U\}$ is finite for every nbhd $U$ of $p.$ 
BTW in a metrizable space $Y,$  a sequence $(f(x))_{x\in \mathbb N}$ converges to $p$ iff $\{p\}=\cap_{y\in \mathbb N}\;Cl(\;\{f(x):x\geq y\}\;).$ In other kinds of spaces this may not be true. For the space $X$ in your Q , let $f:\mathbb N\to \mathbb N^2$ be a bijection.  Then $\{(0,0)\}=\cap_{y\in \mathbb N}Cl(\{f(x):x\geq y\})$ but $(0,0)$ has a nbhd $U$ such that  $\{x\in \mathbb N: f(x)\not \in U\}$ is infinite.
