Accumulation points in $R^2$ and general $R^n$ So I am solving at the moment question of this form first if we take this question. Determine the accumulation point in $R^2$ and if they are closed sets ,open sets, or neither.
All complex numbers z such that $|z| > 1$ 
$S =  \{z \in \mathbb{C} : |z| >  1 \} = \{z = x + iy: x^2 + y^2 > 1 \}$
So according to intuition I can easily see that S` which is the set of accumulation points is the following:
$S` = \{ Z \in \mathbb{C}: |z| \geq 1\}$, however I can't seem to prove it rigorously in terms of the euclidean metric as computations are weird maybe I am not familiar with it yet. I can solve such question in the easy euclidean metric in $R^1$ but $R^n$ metric makes the computations much harder. If someone could show me the technique that would be great.
In the following question above we need to prove that $S`$ is the only accumulation points that is if we consider any ball $B(z`,r)$ such that $z` \in S`$ then $(B(z`,r) - {z`})\cap S` \neq \emptyset$ for any r > 0 and any set outside $S`$ we can always find a counter example.
Edit:
So I was able to prove it first of, however I am having troubles proving to myself outside of $S'$ it will be not accumulation points.
Suppose $S' = \{ z' \in \mathbb{C}: |z'| > 1 \} = \{z' = x' + iy' : x'^2 + y'^2 > 1 \}$
Now consider arbitrarily $r > 0$ and consider $z \in B(z',r)$, where z' is arbitrarily element in $S`$. 
\begin{align} 
d(z',z) < r &\implies (x - x')^2 + (y - y')^2 < r^2 \\
&\implies x^2 + y^2 + x'^2 + y'^2 - 2xx' - 2yy' < r^2 \\
 &\implies x'^2 + y'^2 < r^2 - [x^2 + y^2] + 2xx' + 2yy' 
\end{align} 
So we have 
\begin{align} 
1 < x'^2 + y'^2 < r^2 - [x^2 + y^2] + 2xx' + 2yy' \implies
1 < x^2 + y^2 < r^2 - [x`^2 + y`^2] + 2xx` + 2yy`
\end{align} 
However, we know that $x'^2 + y'^2 > 1$ along with inequality above we have $x^2 + y^2 > 1$, so we must have that for any z $\in B(z',r)$ and any r > 0 we have $(B(z',r) - {z'})\cap S \neq \emptyset$. 
However I am trying to show myself that for $S'^c$, then we won't have any points, @ColdNumber I see it intuitively however I can't prove it rigourously. 
 A: Let's prove the statement for $\Bbb R^n$ (the process is completely analogous for $\Bbb R^2$):
Let $x=(x_1,\ldots,x_n)$ be an arbitrary point of $S'$ and let $\delta$ be an arbitrary positive real number.
For each $k\in\{1,\ldots,n\},$ let $r_k$ be a rational number that satisfies $x_k<r_k<x_k+\frac{\delta}{\sqrt n}$ if $x_k\geq0$ or $x_k>r_k>x_k-\frac{\delta}{\sqrt n}$ if $x_k<0.$ 
It is easy to see that if $r=(r_1,\ldots,r_n),$ then $\|r\|>1$ and $\|r-x\|<\delta,$ so that $(B(x,\delta)-\{x\})\cap S$ is not empty, which proves that $S'$ is a subset of the set of all accumulation points of $S.$
To prove that $S'$ is the set of all accumulation points of $S,$ let $y\in\Bbb R^n-S'.$ Then $\|y\|<1.$
Let $\varepsilon:=\min\{\|y\|,1-\|y\|\}$ (We do not consider the case when $\|y\|=0$ for obvious reasons). Then $B(y,\varepsilon)\subseteq\Bbb R^n-S'.$
Therefore $S'$ is the set of all accumulation points of $S.$
A: I'll address the accumulation points question.
Points whose modulus is strictly greater than  1 will be interior points, so they are automatically accumulation points. To prove this, you could say that if $|z|>1$, then $|z|=1+\varepsilon$ for some $\varepsilon >0$, so $B(z, \frac \varepsilon 2)\subseteq S$, so $z$ is an interior point of $S$.
Similarly, points $z$  whose modulus is strictly less than 1, will have $|z|=1-\varepsilon$, so the open ball $B(z, \frac \varepsilon 2) \subseteq \Bbb C\setminus S$, meaning the ball contains no points of $S$ and hence $z$ cannot be an accumulation point of $S$.
For points on the boundary $|z|=1$:
You can represent these as $e^{i\theta}$, where $\theta$ is the angle the ray from the origin through the point makes with the real axis. (I'm thinking of the complex plane.)
Then you can show that a point $z=e^{i\theta}$ in the boundary will be the limit point of the sequence $\{(1+ 1/n)e^{i\theta}\}$, all of whose terms are in $S$ because their modulus is $1+1/n>1$ , and this will show it is an accumulation point of $S$ (because it shows that any open ball around $z$ will contain infinitely many points of the sequence and hence infinitely many points of $S$.)

EDIT: I will expand on why a point $z$ with $\left| z\right|<1$ cannot be an accumulation point of $S$. 
To show that $z$ is not an accumulation point, we just need to find an open ball around $z$ that contains no points of $S$. 
If a point $z$ has modulus less than $1$, meaning $\left| z\right|<1$, for some positive number $\varepsilon$ we have $\left| z\right| =1-\varepsilon$.
Consider the open ball  $B(z, \varepsilon)$.
If $z' \in B(z,\varepsilon)$, then $\varepsilon >\left| z'-z\right|$. By a variation of the triangle inequality, $\left|z'-z\right|\geq \left|z'\right|-\left|z\right|$, so we have the following:
\begin{align}
 \varepsilon  &> \left|z'-z\right|\geq \left|z'\right|-\left| z\right|=\left| z'\right|-(1-\varepsilon)\\
\varepsilon  &> \left| z'\right|-(1-\varepsilon)\\
(1-\varepsilon)+\varepsilon &> \left| z'\right| \\
1&>\left| z'\right|
\end{align}
We have shown that any point in the open ball $B(z,\varepsilon)$ is not in $S$, which means that $B(z, \varepsilon)\cap S = \varnothing$, so $z$ is not an accumulation point of $S$.
