Question: Suppose a set $K\subset \mathbb{R}$. Every sequence in K has a convergent subsequence that converges to a limit in K. Prove that K is bounded.

I want to use contradiction. Suppose K is not bounded, then there should be an unbounded sequence $\{a\_n\}$. And I want to show that $\{a\_n\}$ doesn't have a convergent subsequence, but I don't know how to do.

Answer: 
* If $x = 0$ then it's obvious that $-|x| \leq x \leq |x|$ because $$-|x| = x = |x| = 0.$$
* If $x > 0$ then $x = |x|$ and
$$ -\underbrace{x}\_{=|x|} < 0 < x =|x|.$$
Therefore $-|x| \leq x \leq |x|.$
* If $x < 0$ then $x = -|x|$ and
$$ -|x| = x< 0 < -\underbrace{x}\_{-|x|} = |x|.$$
Therefore $-|x| \leq x \leq |x|.$

> 
> Why does the distance $x$ has the possibility to be greater than $x$ itself, shouldn't it only be that $x=|x|$ in any case? why greater than?
> 
> 
> 

If $x = y$ than it is also true that $x \leq y$. The latter statement is less precise than the former, sure, but it is a true statement.

At least one of these inequalities 
$$-|x| \leq x \leq |x|$$
is actually an equality. Which one of them ? Well it depends on the nature of $x$ (positive, negative, zero). Therefore it's better to put inequalities because it is true for all $x$.

> 
> I've seen only two different approaches, one that it seems to me too trivial [...] Is that reasoning even valid?
> 
> 
> 

Yes, it is.
The value of $x$ must be either $-|x|$ or $|x|$ depending on the sign of $x$. This can be expressed as the inequality you listed $-|x|\leq x\leq|x|$. I think you are trying to compare it to an inequality like $0\leq x\leq4$ where $x$ could take any value from $0$ to $4$. In your situation this is not the case, $x$ can only be one of two specific values, namely $-|x|$ or $|x|$.

> 
> Why does the distance $x$ has the possibility to be greater than $x$ itself, shouldn't it only be that $x=|x|$ in any case? why greater than?
> 
> 
> 

The distance $|x|$ will always be a non-negative number where as $x$ can be a negative number so in some cases (i.e. $x<0$) we have that the distance is greater than the value (where greater is meaning signed magnitude not just magnitude).