How to find all cluster/accumulation points of a given sequence? 
Let U={$x_n$=($\frac{1}{n}$,(-1)$^n\frac{n-1}{n}$) :n$\in\mathbb{N}$
  }.Find all cluster points of the set U and prove rigoriously.

First of all, the definition of a cluster point: 

x$\in\mathbb{R}^n$ is a cluster/accumulation point of set U$\subset\mathbb{R}^n$ if every
  punctured neighbourhood of x B'(x, r) contains elements from the set U.

What I've done:
Let's take a look at the components $\frac{1}{n}$ and (-1)$^n\frac{n-1}{n}$ separately. It is quite obvious that $\lim_{n\to_\infty} \frac{1}{n}=0$ and that $(-1)^n\frac{n-1}{n}$ diverges. In addition, its subsequences $(-1)\frac{n-1}{n}=\frac{1}{n}-1$ and $\frac{n-1}{n}=1-\frac{1}{n}$  converge to -1 and 1 respectively. Thus my claim is that the cluster points of set U are points (0,1) and (0,-1). Next it needs to be proven in a formal way. 
Since  ($\frac{1}{n}$, $1-\frac{1}{n}$) converges to (0, 1), $\forall\epsilon>0, \exists$M such that ||$x_n$-(0,1)||<$\epsilon$, when n>M.
On the other hand, since  ($\frac{1}{n}$, $1-\frac{1}{n}$) converges to (0, 1), for every neighbourhood U of (0,1) there exists N$\in\mathbb{N}$ such that $x_n$$\in$U whenever n$\geq$N. This implies directly(?) that (0,1) is a cluster point of U (?).  
Similarly, since  ($\frac{1}{n}$, $\frac{1}{n}-1$) converges to (0, -1), for every neighbourhood U of (0,-1) there exists N$\in\mathbb{N}$ such that $x_n$$\in$U whenever n$\geq$N..?
Lastly, it must be shown that (0,-1) and (0,1) are the only cluster points of the set U. By contradiction?
 A: Here we show that $(0,1)$ and $(0,-1)$ are the only cluster points of the set
\begin{align*}
U=\left\{x_n=\left(\frac{1}{n},(-1)^n\left(1-\frac{1}{n}\right)\right):n\in\mathbb{N}\right\}
\end{align*}

We argue by contradiction: Assuming $P=(x,y)$ different to $(0,1)$ and $(0,-1)$ is a cluster point, we prove there is a punctured disc 
  $B(P,r)$ which does not contain a point of $U$, thus violating the claim of $P$ being a cluster point.

At first we take a look at the points $x_n$ of $U$. We observe, that for each $\varepsilon>0$ the points
\begin{align*}
x_{2n}&=\left(\frac{1}{2n},1-\frac{1}{2n}\right)\in B((0,1),\varepsilon)\qquad\qquad &2n\geq \frac{1}{\varepsilon}\\
\text{and}\qquad\qquad &\\
x_{2n+1}&=\left(\frac{1}{2n+1},-1+\frac{1}{2n+1}\right)\in B((0,-1),\varepsilon)\qquad\qquad &2n+1\geq \frac{1}{\varepsilon}
 \end{align*}
We observe: For each $\varepsilon>0$ there is $N(\varepsilon)\in\mathbb{N}$  so that for all $n>N(\varepsilon)$ the point $x_n$ is either element of $B((0,1),\varepsilon)$ or $B((0,-1),\varepsilon)$.

Now assume a point $P=(x,y)$ different to $(0,1)$ and $(0,-1)$ is given. We consider a punctured disc $B(P,r)$ with center $P$ and radius $r$ less than the minimum distance to $(0,1)$ and $(0,-1)$.
We choose $\varepsilon>0$ so that
  \begin{align*}
B(P,r)\cap B((0,1),\varepsilon)=\emptyset\qquad\text{and}\qquad B(P,r)\cap B((0,-1),\varepsilon)=\emptyset
\end{align*}
According to the arguments above we can find $N(\varepsilon)\in\mathbb{N}$ so that
$$x_n\in \left(B((0,1),\varepsilon\right)\cup B((0,-1),\varepsilon)\qquad\qquad\qquad n\geq N(\varepsilon)$$
This way it is guaranteed that at most finitely many members $x_n, 1\leq n<N(\varepsilon)$ can be elements of the punctured disc $B(P,r)$.

We now shrink the punctured disc to also exclude these finitely many points $x_n, 1\leq n<N(\varepsilon)$.

We consider a radius $r_{N(\varepsilon)}$ less than the minimum positive distance from the center $P$ to the points $\{x_1,\ldots,x_N(\varepsilon)\}$ and less than $r$. The so constructed punctured disc contains no members of $U$
  \begin{align*}
B(P,r_{N(\varepsilon)})\cap U=\emptyset
\end{align*}
  showing that $P$ is not a cluster point.

                                                       
Note: When requiring $r_{N(\varepsilon)}$ to be less than the minimum positive distance we also respect a situation whereby one point of $U$ is  $P$.
