Infinite subset of $\mathbb{R}$ with exactly 4 limit points: any common example? 
Give an example of an infinite set $S\subset \mathbb{R}$ such that S has
   1) only 1 limit point
   2) exactly 4 limit points

In case of (1), the infinite set $S=\{\frac{1}{n}:n\in\mathbb{N}\}\subset\mathbb{R}$ is a commonly used example.
My solution to (2):
Define sets $A,B,C,D\subset\mathbb{R}\backepsilon$
$A=\{\frac{1}{n}:n\in\mathbb{N}\}\Rightarrow A'=\{0\}$ ,
$B=\{1+\frac{1}{n}:n\in\mathbb{N}\} \Rightarrow B'=\{1\}$ ,
$C=\{2+\frac{1}{n}:n\in\mathbb{N}\} \Rightarrow C'=\{2\}$ , and
$D=\{3+\frac{1}{n}:n\in\mathbb{N}\} \Rightarrow D'=\{3\}$
$S:=A\cup B\cup C\cup D\subset\mathbb{R}$ is an infinite set amd
$S'=(A\cup B\cup C\cup D)'=A'\cup B'\cup C'\cup D'=\{0,1,2,3\}$

My question:Is there a more general (or commonly used, as in case 1) solution to (2)?

 A: Choose four non-eventually constant convergent sequences that converge to four different points, and define a set by the union of the elements of these sequences.
Also you can proceed choosing first four different points of the real line and join together in a set any finite number of non-eventually constant sequences that converge to these points.
A: Let's define $\displaystyle f(n)=\sin\left(\frac{\pi}3+\frac{n\pi}{2}+\frac 1{n}\right)$
Then $S=f(\mathbb N)$ has $4$ limit points 
$\sin(\frac\pi 3)=\frac{\sqrt{3}}2,\ \sin(\frac{5\pi}6)=\frac 12,\ \sin(\frac{4\pi}3)=-\frac{\sqrt{3}}2,\ \sin(\frac{11\pi}6)=-\frac 12$

Edit: addendum regarding your question in comment.
In fact as long as you have a continuous periodic function $f$ of period $T$ you can easily build limit points.
Remember that we need $4$ convergent sequences so let's take a random starting point $x_0$ and set:
$\begin{cases}
\ell_a=f(x_0)&=f(x_0+nT)\\
\ell_b=f(x_0+\frac T4)&=f(x_0+\frac T4+nT)\\
\ell_c=f(x_0+\frac{2T}4)&=f(x_0+\frac{2T}4+nT)\\
\ell_d=f(x_0+\frac{3T}4)&=f(x_0+\frac{3T}4+nT)\\
\end{cases}$
You just need to choose $x_0$ carefully such that the four $\ell_i$ are different.
This is why I had chosen $x_0=\frac{\pi}3$ in my previous example. For instance if you take $x_0=0$ or $x_0=\frac{\pi}4$ because of the symmetries of $\sin$ you have some values which are equal.
Of course if you want more than $4$ limits, just take $\frac Tm$ where $m$ is the number of limit points you want.
Notice now that we can regroup our four limits into a cycling sequence: $u_n=f(x_0+\frac{nT}4)$
But this sequence takes only $4$ different values $\ell_a,\ell_b,\ell_c,\ell_d$, so in order to have accumulation point we need to add a small variation around these points. 
For instance $v_n=f(x_0+\frac {nT}4+\frac 1n)$ works, but $w_n=f(x_0+\frac {nT}4)+\frac 1{n^2}$ works as well.
Since $f$ is continuous, it does not matter whether you add the epsilon term inside the function or outside.
You can build many examples on this simple model!
