Finding $\left(\bigcup_{n\in\Bbb N}S_n\right)', S_n=\{(x,y)\in\Bbb R^2\mid \|(x,y)-(1/n,1/n)\|_2=1/n,xy\ne 0\}$ 
Let $S_n=\left\{(x,y)\in\Bbb R^2\mid \left\|(x,y)-\left(\frac1n,\frac1n\right)\right\|_2=\frac1n,xy\ne 0\right\}$ and let $S=\bigcup_{n\in\Bbb N} S_n.$ Find the set $S'$ of limit points of $S$.

My answer which I would like to verify:

I claim $S'=S\cup\bigcup\limits_{n\in\Bbb N}\left\{\left(\frac1n,0\right),\left(0,\frac1n\right)\right\}\cup\{(0,0)\}.$

Let $A_n=\left\{(x,y)\in\Bbb R^2\mid\left\|(x,y)-\left(\frac1n,\frac1n\right)\right\|_2=\frac1n\right\}$ and $(x_0,y_0)\in A_n$ arbitrary. I want to show that $$\forall r>0, B((x_0,y_0),r)\cap A_n\setminus\{(x_0,y_0)\}\ne\emptyset.$$
Let $r>0$ be arbitrary. According to the Archimedean axiom, we can find $m_1,m_2\in\Bbb N$ s. t. $\frac1{m_1}<r,\frac1{m_2}<\frac2n$. Let's take $m=\max\{m_1,m_2\}.$ Now, the circle $B_{1m}$ of the radius $\frac1m$ centered at $(x_0,y_0)$ and the circle $A_n$ are intersecting each other at two points $P_1$ and $P_2$. Then $P_1$ and $P_2$ are also in the open ball $B((x_0,y_0,r).$ Since this holds for any $r>0,$ we obtain a sequence of infinitely many different points converging to $(x_0,y_0)$. Therefore, $(x_0,y_0)\in S'.$
Now, let's consider the sequence of the intersection points $(Q_n)_n$ of the line $y=x$ and the circles $(A_n)_n.$ Then, $(Q_n)_n$ is a sequence of infinitely many different points converging to the origin, therefore $(0,0)\in S'$. Now, let's show the rest of the points in $\Bbb R^2$ cannot be limit points of $S$. I'm going to use the following notation:
$E=\{(x,y)\in\Bbb R^2\mid\|(x,y)-(1,1)\|_2>1\}\cap\{(x,y)\in\Bbb R^2\mid x<0\text{ or } x>1\text{ or } y<0\text{ or } y>1\},$
$F_n=\left\{(x,y)\in\Bbb R^2\mid\left\|(x,y)-\left(\frac1n,\frac1n\right)\right\|_2<\frac1n,\left\|(x,y)-\left(\frac1{n+1},\frac1{n+1}\right)\right\|_2>\frac1{n+1}\right\},n\in\Bbb N$
$G_n=\left\{(x,y)\in\Bbb R^2\mid\left\|(x,y)-\left(\frac1n,\frac1n\right)\right\|_2>\frac1n,\left\|(x,y)-\left(\frac1{n+1},\frac1{n+1}\right)\right\|_2>\frac1{n+1}\right\},n\in\Bbb N$
$H_n=\left\{(x,y)\in\Bbb R^2\mid \frac1{n+1}<x<\frac1n\right\}\cup\left\{(x,y)\in\Bbb R^2\mid \frac1{n+1}<y<\frac1n\right\},n\in\Bbb N$
All of the above sets are open. Hence, $I_n:=F_n\cup G_n\cap H_n$ is open $\forall n\in\Bbb N$
The remaining points are in the set $ J=E\cup\bigcup\limits_{n\in\Bbb N} I_n,$ which is open. It follows that $\Bbb R^2\setminus J$ is a closed set containing $S,$ which means $\Bbb R^2\setminus J\supseteq\overline S.$ This was sufficient to prove all the investigated $(x_0,y_0)$'s are the only limit points. Alternatively, since $J$ is open, for each $(x_0,y_0)$ we could've found $r_x>0$ s. t. $B((x_0,y_0),r_x)\subseteq J\subseteq\Bbb R^2\setminus S,$ meaning there is an open neighbourhood of $(x_0,y_0)\in J$ in which there is no point from $S$.
I also found a similar question with the norm $\|\cdot\|_\infty,$ so I would also like to ask if the procedure can be generalized to an arbitrary $p$ norm, $1\le p<\infty.$
 A: I agree that
$$S'=S\cup\bigcup_{n=1}^\infty\{(1/n,0),(0,1/n)\}\cup\{0\}, \tag{1}$$
but I propose a simpler proof. One thing worth noting is that the right-hand side of $(1)$ can be written as $\{0\}\cup\bigcup_{n=1}^\infty A_n$, where $A_n$ is the sphere with center $(1/n,1/n)$ and radius $1/n$ (as defined in the question).
I claim that the inclusion "$\supseteq$" is obvious since it should be clear how to come up with approximating sequences for all the proposed limit points.
For the other inclusion, take $x\in S'$. If $\|x\|=0$, then $x=0$ so it is clearly an element in the right-hand side of $(1)$. If $\|x\|>0$ we let $(x_k)\subseteq S$ be a sequence converging to $x$. Then $\|x_k\|>\|x\|/2$ for all $k$ large enough. This implies that there exists $n_0\in\mathbb{N}$ such that $x_k\in\bigcup_{n=1}^{n_0} A_n$ for all $k$ large enough. Since $\bigcup_{n=1}^{n_0} A_n$ is compact it follows that $x\in\bigcup_{n=1}^{n_0} A_n$. This shows that $x$ is an element in the right-hand side of $(1)$.
