Does every sequence have at least one limit point? Let $x_n$ be a sequence of real numbers.
Definition: $x \in \mathbb R \cup \{-\infty,\infty\}$ is a limit point of a sequence $x_n$  if there is a subsequence  $x_{n_k}$ of our sequence such that $x_{n_k} \to x$. 

If $A= \{x \in \mathbb R \cup \{-\infty,\infty\} \mid x \text { is a limit point of }x_n\}$, prove that $A$ is not empty.

 A: If the sequence is bounded, then you can invoke Bolzano-Weierstrass theorem.
So we can assume it is unbounded. Let us assume it is not bounded from above. (The symmetric case is similar.)
Thus for every $C$ there exists $n$ such that $x_n>C$.
Start with $C_1=1$ and choose $n_1$ such that $x_{n_1}>C_1=1$.
Now choose $C_2=2+\max\{x_j\; : \; j\le n_1\}$. Obviously $C_2\ge 2$ and you can choose $n_2$ such that $n_2>n_1$ and $x_{n_2}>C_2$. 
You continue by induction. In the $k$-th step you choose $C_k=k+\max\{x_j \; : \; j\le n_k\}$. This means that $C_k\ge k$ and you there exists $n_{k+1}>n_k$ with $x_{n_k}>C_k$.
In this way you obtain a subsequence $n_k$ such that $x_{n_k} \ge k$. This implies that this subsequence converges to $+\infty$.
A: $\arctan : [-\infty,\infty] \rightarrow [-\pi/2,\pi/2]$.  The sequence $\{x_n\}_{n=1}^\infty \subseteq \mathbb{R}$ has a limit point in $[-\infty,\infty]$ if and only if the sequence $\{\arctan x_n\}_{n=1}^\infty \subseteq [-\pi/2,\pi/2]$ has a limit point in $[-\pi/2,\pi/2]$.  Bolzano-Weierstrass takes care of that.
A: A proof sketch assuming Bolzano-Weierstrass. 


*

*Show that $x_n$ is bounded above if and only if $\infty$ is not a limit point of $x_n$.

*Reversing the above argument, show that $x_n$ is bounded from below if and only if $- \infty$ is not a limit point.

*From (1.) and (2.), conclude that if neither $\infty$ nor $-\infty$ is a limit point of $x_n$, then $x_n$ is bounded. Further, using Bolzano-Weierstrass, conclude that if neither $\infty$ nor $-\infty$ is a limit point of $x_n$, then $x_n$ has a subsequence converging to some limit $x \in \mathbb R$. 
To summarize the argument, the key idea is that the sequence is 


*

*either unbounded, in which case one of $+\infty$ and $-\infty$ is a limit point;

*or it's bounded, in which case Bolzano-Weierstrass shows the existence of a limit point in $\mathbb R$. 


Note: Martin's answer explains the idea behind (1.).  :-)
A: Define $y_n = \sup \{ x_n, x_{n+1}, x_{n+2}, \cdots \} .$ Then $y_n$ is a monotone decreasing sequence and hence $ Y = \displaystyle \lim_{n\to\infty} y_n$ exists (possibly $-\infty$). 
The claim is that $ Y \in A .$ If $Y$ is finite, then argue as follow: For any $\epsilon > 0 $, by the definition of supremum for each $n\in \mathbb{N}$ we can find $ n_k \geq n $ such that $ |y_n - x_{n_k} | < \frac{\epsilon}{2}.$ By definition of limit, for some $n_0 \in\mathbb{N}$ we have $ |Y - y_n| < \frac{\epsilon}{2} $ for all $ n > n_0.$ Thus for all $n> n_0$, $$ |Y - x_{n_k}| \leq |Y-y_n| + |y_n - x_{n_k}| < \epsilon.$$ Thus, $ x_{n_k} \to Y.$
I leave the case where $ Y= -\infty$ to you. 
