Compact set is closed and bounded - correct idea? 
Prove that a compact set $A\subseteq\mathbb{R}^n$ is closed and bounded.

My attempt:
I have already shown that $A$ must be bounded.
To show that $A$ is closed I want to show that  $\mathbb{R}^n\setminus A$ is open.
For that, let $x\in \mathbb{R}^n\setminus A$ be arbitrary. Suppose now, for  the sake
of contradiction, that there didn't exists any $\epsilon>0$ s.t. $B_{\epsilon}(x) 
\subseteq \mathbb{R}^n\setminus A$. Then we can consider the sequence
$(\epsilon_j)_{j\in\mathbb{N}}$ s.t. for all $j\in \mathbb{N}$
$\epsilon_{j} >0$  and $\epsilon_{j+1}<\epsilon_{j}$. For each $\epsilon_{j} $ there must
exist some $y_{j} $ s.t. $y_{j} \in A \cap B_{\epsilon_{j} }(x) $. Now I tried to construct an infinite open cover of the points in $Y=\{y_{j}  \mid j\in \mathbb{N}\} $:
$$
\mathcal{O}(Y)=\{(y_{j} -f_{j} , y_{j} +f_{j}  ) \mid j\in \mathbb{N} \text{ and }
f_{j} = \min_{\large y_{k }\in \{y_{i} \  \mid\  i\in \mathbb{N} \}\setminus \{y_{j}\}   }
\left\|y_{j} -y_{k} \right\| \}
.
$$
The reason why I define the cover in such a  weird way is that I want to cover all the $y_j$'s but don't want a finite number of open sets to  overlap in such a way that they cover all $y_j$; however, I'm very unsure whether my attempt in doing this was successful.
Then I want some cover $\mathcal{V}$ which covers  all yet uncovered
points but doens't cover any $y_{j} $. Next, when considering the cover
$$
\mathcal{O}(A) =  \mathcal{O}(Y)\cup \mathcal{V}
.$$
which doesn't have any finite subcover that covers $A$ (because of the $y_j$'s) which contradicts the
fact that $A$ is compact. Is my idea correct (meaning that I can make it rigorous)?
Note: $B_{\epsilon}(x) =\{y\in \mathbb{R}^n \mid \left\|y-x\right\| <\epsilon\}$
 A: Your problem is you are putting the $y_i$ close to $x$ first and trying to build on open cover on just them and not on all the points in $A$.
Find an open cover for all $w\in A$ via $B_{r_w}(w)$.  And we can "keep" all these neighborhoods away from $x$ by letting $r_w =\frac 12 d(w,x)$.  That way for all the points $y \in B_{r_w}(w,x)$ we have $d(y,x) > r_w$.
And that's good.  As $\{B_{r_w}(w)\}$ is an open cover and $A$ is compact we have finite subcover $U$.  If we consider the values of the $r_\alpha$ radii of the $B_{r_\alpha}(\alpha)$ sets in the finite subcover, there are a finite number of them so there is a least radius $r$.
Now as all $w \in A$ are in some $B_{r_y}(y)$ in the finite subcover (not nesc. centered at $w$) and as $r_y \ge r$ then all $w \in A$ are so that $d(w,x) > r$.
And that's it.... if $\epsilon = r = \min r_\alpha = \min \{r_\alpha|$ indexes of the finite subcover of the $B_{r_\alpha}(\alpha)\}$ then if $d(x, u) < r$ then $u\not \in A$.  So $B_r(x) \subset \mathbb R^n \setminus A$.
So $\mathbb R^n\setminus A$ is open.
.... still not the most straightforward way to go.
