Bounded sets are relatively compact in $\mathbb{R}$ I know that every closed and bounded set in $\mathbb{R}$ is compact (like $[a,b]$)
so i can conclude that every bounded set in $\mathbb{R}$ is relatively compact, by contradiction i say that let $A\subset\mathbb{R}$ be a bounded set but not relatively compact it means that $\overline{A}$ is not compact, but $\overline{A}$ is closed and bounded in $\mathbb{R}$ so it is compact, contradiction. 
My question is if $A$ is bounded, why $\overline{A}$ still bounded ?
Thank you.

If $A$ is bounded this means there is some $R > 0$ such that $d(x,y) \le R$ for all $x,y \in  A$. But the same then holds for $\overline{A}$ as well. This can for instance be seen as follows: pick $p \in A$, then all members of $A$ are in the closed ball $D(p,R) = \{x \in X: d(x,p) \le R \}$. So $\overline{A} \subseteq \overline{D(p,R)} = D(p,R)$ as well.
So If $A$ is bounded, $\overline{A}$ is bounded (see above) and closed (by definition), so if we are in $\mathbb{R}$ or $\mathbb{R}^n$, $\overline{A}$ is compact by Heine-Borel.
