sequence in $\mathbb{N}\times\mathbb{N}\cup\{(0,0)\}$ with $(0,0)$ as accumulation point 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  there is a sequence in $X\setminus\{(0,0)\}$ for which $(0,0)$ is an accumulation point but no subsequence converges to $\{(0,0)\}$.  
At first I though that for every neighbourhood of $(0,0)$ I could select an $m$ such that $T:=\{n\in\mathbb{N}\cap \{0\}\mid (n,m)\not\in U\}$ is finite and set $x_m=(n_0,m)\in T$ for a fixed $n_0\in\mathbb{N}$, but I there might be uncountable many neighbourhood of $\{(0,0)\}$, so this does not work. Can anybody help me? 
 A: How about the standard diagonal sequence that is used to show $\mathbb{N}\times\mathbb{N}$ is countable? I.e. $(a_n)=((1,1),(1,2),(2,1),(3,1),(2,2),(1,3),\dots)$? For any neighborhood of $U$, we have a nonempty intersection $$U\cap\left(\cup_{n}\{a_n\}\right)$$
Now choose a subsequence $(a_{n_k})=(s_{n_k},t_{n_k})$. There are a few possibilities: 
Case 1: the set $\cup_{k}\{t_{n_k}\}$ is finite. In this case, there is some $m$ such that $t_{n_k}=m$ infinitely often. Letting $$U=\left(\mathbb{N}\times\mathbb{N}\right)\setminus\{(n,m):n\in\mathbb{N}\},$$
we can see that this $U$ is a neighborhood of $(0,0)$ for which $a_{n_k}\not\in U$ infinitely often, so this subsequence does not converge to $(0,0)$.
Case 2: the set $\cup_{k}\{t_{n_k}\}$ is infinite. In that case, using the axiom of choice, choose ordered pairs from the subsequence $V=\{(s_{n_k},t_{n_k})\}$ so that each ordered pair has a unique $t$ element. (So for each $t$ that appears in an ordered pair in the subsequence, simply choose one ordered pair that it appears in). Now consider 
$$U=\left(\mathbb{N}\times\mathbb{N}\right)\setminus V.$$
What can you say about $U$ and it's intersection with your subsequence?
