proving a sequence has exactly four limits points Given the sequence \begin{align*}
x_n:=\begin{cases}
1,&\textrm{if }n\equiv0 \mod 4\\
2,&\textrm{if }n\equiv 1\mod 4\\
3,&\textrm{if }n\equiv 2\mod 4\\
4,&\textrm{if }n\equiv 3\mod4
\end{cases}
\end{align*}
I want to show that the limit points are exactly $1,2,3,4$
So $1$ is a limit point since $x_{4n}$ converges to $1$.
$2$ is a limit point since $x_{4n+1}$ converges to $2$.
Doing the same for $3,4$ I know $1,2,3,4$ are limit points.
But why are there no other limit points?
Suppose $y$ is another limits, $y\neq1,2,3,4$. So there is a subsequence $x_{n_l}$ with $x_{n_l}\rightarrow y$ for $l\rightarrow \infty$.
Now I am stuck. How can you get a contradiction?
 A: Choose $\epsilon = \frac{1}{2}\min\{|y-1|,|y-2|,|y-3|,|y-4|\}$ (so e.g., if $y = 1.1$, take $\epsilon = 0.05$.)
Then you know that for every $n$, $|x_n - y| > \epsilon$. Do you see how you get a contradiction now? 
Full proof:
Suppose $y$ is a limit point of this sequence. Then, for every $\epsilon > 0$ there is a subsequence $\{x_{n_l}\}$ such that there exists $L$ so that whenever $l > L$, we have $|x_{n_l} - y| < \epsilon.$
Now, choose $\epsilon = \frac{1}{2}\min\{|y-1|,|y-2|,|y-3|,|y-4|\}$. If $y$ is not $1,2,3$ or $4$, then $\epsilon > 0$, and we have for every $x_n$ (i.e. not just the subsequence) that $|x_n - y| > \epsilon$. Thus, there can be no such $L$ as described in the first paragraph; i.e. we cannot simulaneously have 
$$
|x_{n_l} - y| < \epsilon
$$
as required for convergence as well as 
$$
|x_{n_1} - y| > \epsilon
$$
as we just demonstrated. Thus we have a contradiction.
A: Write out the definition of convergence of a sequence (for each $\delta$ there is an $\epsilon$...), negate it (there is a $\delta$ such that for all $\epsilon$...) and show an example of that.
A: Forget about the sequence itself. Just try to find the limit points of the set $\{ 1,2,3,4\}$. We are looking for all sequences in $\{ 1,2,3,4\}$ which converge to some $x \in \mathbb R$. The definition of convergence involves an $\varepsilon$ and $N$. What happens if we choose $\varepsilon = 1/2$?
A: Following your idea, if $x_{n_k}\to y$ then there is an $\ell<4$ so that there are infinitely many $k$ such that $n_k\equiv \ell\  \rm{mod(4)}$. It follows that $\lim x_{n_k}=\ell=y$ which is a contradiction.
