A local base problem The following question is an example(without proof) in my lecture notes. I'm trying to figure out the proof.

Consider the subset of $\mathbb{R}^2$:
$$X = \{(0,0)\}\cup\{(2^{-n},0):n \in \mathbb{N}\}\cup\{(2^{-n},2^{-m}):n，m \in \mathbb{N}\}.$$
The points different from $(0,0)$ are assigned to their natural beighborhoods.
  But at $(0,0)$, for every $n\in \mathbb{N}$ and every function $f:\mathbb{N}\rightarrow\mathbb{N}$,
  we set
$$B(f,n) = \{(0,0)\}\cup\{(2^{-m},0):m \geq n\}\cup\{(2^{-m},2^{-l}):n，m \geq n, l\geq f(m)\}$$, then set
  $\mathcal{B}_{(0,0)} = \{B(f,n)\}_{f,n}$

Claim:


*

*the assignment is a valid assignment of local bases.

*$(0,0)$ is in the closure of $A =   
    \{(2^{-n},2^{-m}):n,m\in\mathbb{N}\}$ but no sequence from A 
converges  to $(0,0)$.


Since it is discrete (apart from $(0,0)$), every point is open set, so the claim that natural neighborhoods of points not $(0,0)$ are local bases is trivial.
But I dont know how to argue about $(0,0)$. furthermore, should I use the result from claim 1 to prove claim 2?
Thank you in advance. 
 A: I assume that by natural neighborhoods you mean the open neighborhoods in the relative topology inherited from the usual topology of $\mathbb{R}^2$. In other words, the local basis at $x\in X,x\neq(0,0)$ is $\mathcal{B}_x=\{X\cap U:U\text{ is open in } \mathbb{R}^2,x\in U\}$.
In order to be a system of local bases for a topology of $X$ the set $\{\mathcal{B}_x:x\in X\}$ has to have the following three properties:
$$
\begin{align*}
&(1)\quad \mathcal{B}_x\neq\emptyset\text{ for every }x\in X;\ x\in U\text{ for every } U\in\mathcal{B}_x, \\
&(2)\quad \text{For every }x\in X,\ U,V\in\mathcal{B}_x\text{ there is a } W\in\mathcal{B}_x\text{ with }W\subseteq U\cap V, \\
&(3)\quad \text{For every }x,y\in X,\  x\in U\in\mathcal{B}_y \text{ implies that } V\subseteq U\text{ for some }V\in\mathcal{B}_x.
\end{align*}
$$
The property $(1)$ is clear for every $x\in X$.
The property $(2)$ is clear for every $x\in X\setminus\{(0,0)\}$. For $x=(0,0)$, take $U=B(f,n)$ and $V=B(g,m)$, then $B(\max\{f,g\},\max\{n,m\})\subseteq U\cap V$.
The property $(3)$ is clear if both $x$ and $y$ are in $X\setminus\{(0,0)\}$, so two cases remain. First if $x\in X\setminus\{(0,0)\}$, $y=(0,0)$ and $x\in B(f,n)$, then $B(f,n)\in\mathcal{B}_x$ so choose $V=B(f,n)$. The remaining case is $y\in X\setminus\{(0,0)\}$ and $x=(0,0)\in U\in \mathcal{B}_y$. Note that since $U$ is open in the relative topology, there is an $r>0$ so that $S=\{(x_1,x_2)\in X: 0\leq x_1\leq r, 0\leq x_2\leq r\}\subseteq U$. Now if you take $n\in\mathbb{N}$ with $2^{-n}\leq r$, then $B(\operatorname{id}_\mathbb{N},n)\subseteq S\subseteq U$.
We have shown that $\mathcal{B}_x$'s are a system of local bases for a topology. Note that while all the points of the form $(2^{-n},2^{-m})$ are isolated, the points $(2^{-n},0),n\in\mathbb{N}$ are not: for any neighborhood $U$ of the point $(2^{-n},0)$ there is an $m\in\mathbb{N}$ so that $\{(2^{-n},2^{-l}):l\geq m\}\subseteq U$.
Since $(2^{-n},2^{-f(n)})\in A\cap B(f,n)$ for any $f,n$, the point $(0,0)$ is in the closure of $A$. Assume that $(x_n)$ is a sequence in $A$. Write $x_n=(2^{-f(n)},2^{-g(n)})$. Then $B(g+1,1)$ is a neighborhood of $(0,0)$ containing no elements of $(x_n)$ so $(x_n)$ does not converge to $(0,0)$.
