Proving E satisfies Heine-Borel Property then E is compact May I please receive help proving the following? Thank you
$\def\R{{\mathbb R}}
\def\p{{\bf p}}$ 
Let $E\subseteq\R^n$.
(i)   Prove if $E$ satisfies the Heine-Borel Property, then $E$ is bounded.  
$\textbf{Proof:}$ Consider a compact subset $E\subseteq \R^n$. Now, let $G= \{(-n,n) | n\in \R^n\}$ be an open cover of $E$. However, $E$ is compact. So, $G$ must have a finite subcover as $$G' = \{(-n_1,n_1), (-n_2,n_2), \dots (-n_k, n_k)\}$$ which also covers $E$. 
Let us define a maximum as $m = \max\{(-n_1,n_1), (-n_2,n_2), \dots (-n_k, n_k)\}$. So, $E \subseteq (-m,m)$ and means that the maximum open set in the finite subcover will contain the set $E$. So we can say $|e| < m$ for all $e$ in $E$. Thus, we can say that $E$ is bounded. 
(ii)  Prove if $E$ satisfies the Heine-Borel Property, then $E$ is closed.
$\textbf{Proof:}$ Consider a compact set $E\subseteq\R^n$. Now, assume a point $\p$ that is not in $E$ and instead is in the complement of $E$ as $\p\in E^\complement$. Recall, a point, or a singleton set, can be represented as the intersection of closed balls, or $\displaystyle{\p = \bigcap_{r>0} \overline{B_r} (\p)}$ where $r$ is radius of the closed ball. 
As $E$ is compact, there is a finite subcover $\displaystyle{\bigcup_{a \in F} \mathcal{O}_a}$, for every open cover $\mathcal{O} = \{\mathcal{O}_a | a \in A\}$ in $E$, that covers $E$. So, $\displaystyle{E_k \subseteq \bigcup_{a\in F} \mathcal{O}_a}$, $E$ lies in the finite union of open balls.
Now, consider $\p \in E^\complement$. The singleton set $\{\p\}$ can be represented as the intersection of closed balls, or $\displaystyle{\bigcap_{r>0} \overline{B_r}(p)}$ where $r$ is radius of the closed ball. Use De Morgan's law as $\displaystyle{A^\complement = (\cup B)^\complement = \cap(B^\complement)}$ where $A$ and $B$ are any sets. Thus, $\p \in E$. 
Now, consider $E$. $E$ can be covered by complements of closed balls of $\p$. Let us consider the complements so we get $$E^\complement = \left( \bigcap_{r>0} \overline{B_r}(\p) \right)^\complement = \bigcup_{r>0}\left(\overline{B_r}(\p) \right)^\complement.$$ We know that a complement of a closed set gives an open set. So, it can be seen that $\p^\complement$ is the union of finite open sets. Thus, $E^\complement$ is open. Complement of an open set is closed. So, we get $E$ is a closed set. 
 A: The Heine-Borel theorem is that a set in $\mathbb{R}^N$ is compact iff it is closed and bounded. Presumably you want a proof that if $E$ is compact then it must be bounded, because if $E$ is assumed to be closed and bounded you would be done?
Suppose $E$ is compact but unbounded. If the set is unbounded, there is a sequence $\{x_n\}$ in $E$ that has no convergent subsequence in $E$ (start with an open ball of radius $r_1$ around the origin large enough to include at least one point of $E$ and pick a point of $E$, then pick a point of $E$ from the ball of radius $r_2> r_1$ large enough to include additional points of $E$ complement the ball of radius $r_1$, and so on --- because $E$ is unbounded you will never run out of such balls and this sequence diverges.). But this contradicts the Bolzano-Weierstrass theorem, which is that compact sets in $\mathbb{R}^N$ are sequentially compact, and all sequences in compact sets have convergent subsequences.  Therefore $E$ must be compact.
A: You were close on the closed-part: If $E$ is not closed, it has a limit point $p$ that is not in $E$. Write $\{p\} = \bigcap_n D(p, \frac1n)$ the intersection of closed balls, just as you did.
It follows that $E \subseteq X\setminus \{p\} \subseteq \bigcup_n D(p, \frac1n)^\complement$ by de Morgan, so we have a countable open cover of $E$, which should have a finite subcover $D(p, \frac{1}{n_1})^\complement,\ldots,D(p, \frac{1}{n_k})^\complement$. Let $N > \max(n_1,\ldots, n_k)$ so that $\frac{1}{N}$ is smaller than all used radii in the finite cover. Then, as $p$ is a limit point, we have some $x \in B(p, \frac{1}{N}) \cap E$. But then $x \in B(x, \frac{1}{N}) \subseteq D(p, \frac{1}{n_i})$ for all $i$ and so is not covered by the supposedly finite subcover, contradiction. QED.
As an alternative: let $p \notin E$. Then for each $x \in E$ we take $O_x:=B(x, r_x)$ with $r_x=\frac{d(x,p)}{2} >0$. Then $\{O_x: x \in E\}$ is an open cover of $E$, so has a finite subcover $O_{x_1},\ldots, O_{x_n}$. Then setting $r=\min(r_{x_1},\ldots,r_{x_n})>0$ we find (check this!) that $B(p,r) \cap E = \emptyset$, showing that $E$ is closed, as $p \notin E$ was arbitrary.
For the boundedness, it's easier to take $p \in E$ and as the cover all balls $B(p,n), n \in \Bbb N$ which has a finite subcover. The ball with largest radius contains all of $E$ and so $E$ is bounded (e.g.the diameter is $\le$ two times that radius, or whatever your definition of a bounded subset is). Your notation $(-n,n): n \in \Bbb R^n$ makes no sense. Maybe you meant $(-m,m)^n \subseteq \Bbb R^n$, which would be OK (and $m$ runs to infinity).
