how to show that the compactness of set implies closedness? 
Let $S$ be a subset of $\mathbb R^n $. Then the following statements are equivalent
  a) $S$ is compact
  b) $S$ is closed and bounded
  c) Every infinite subset $S$ has an accumulation point in $S$

I think I terribly misunderstood the compactness and open covering. It's clear that $b) \implies a)$, I am having hard time to understand why $a) \implies b)$, isn't the open interval $(1,3)$ covered by finite open sub-covering of the form $(n,n +1)$.
The proof of my book can be summarized as follows. 
Let $y$ be accumulation point in $S| y \notin S , \forall x \in S,$ choose $r_x = \frac12||x-y||$, then $ \{ B(x, r_x): x \in S\}  $ is open sub-covering of $S$.i.e $S \subseteq \cup_{k=1}^p B(x_k, r_x )$, choosing $r = \frac 12 \min\{r_k\}$, we can show that $\cup_{k=1}^nB(x_k, r) \cap B(y, r) = \emptyset$.
But what's the point? don't we have freedom to choose our $r$? If we define the covering this way, no matter how many open sub-covers we make, we will never cover this the region between $r$ and $y$??
 A: You wrote:

The proof of my book can be summarized as follows.
Let $y$ be accumulation point in $S\mid y \notin S , \forall x \in S,$ choose $r_x = \frac12\|x-y\|$, then $ \{ B(x, r_x): x \in S\}  $ is open sub-covering of $S$.i.e $S \subseteq \cup_{k=1}^p B(x_k, r_x )$, choosing $r = \frac 12 \min\{r_k\}$, we can show that $\cup_{k=1}^nB(x_k, r) \cap B(y, r) = \emptyset$.

Part of the problem is that the proof in the book cannot be summarized that way (unless the book is so badly written that you’d be better off throwing it away!).

*

*$\{B(x,r_x):x\in S\}$ is not an ‘open sub-covering of $S$’; it’s the open cover of $S$ with which you’re starting.


*You never say what $x_1,\dots,x_p$ are.


*The notation $B(x_k,r_k)$ is undefined: either you must write $B(x_k,r_{x_k})$, or for $k=1,\dots,p$ you must define $r_k=r_{x_k}$ and then write $B(x_k,r_k)$.


*There is no need for the factor of $\frac12$ in the definition of $r$ (though I suppose that the author might have added this unnecessary touch).
Here’s a correct version, with a little extra explanatory detail:

Let $y$ be an accumulation point of $S$ that is not in $S$. For each $x\in S$ let $r_x=\frac12\|x-y\|$; since $y\notin S$, $\|x-y\|>0$ for all $x\in S$, and therefore $r_x>0$ for each $x\in S$. Thus, $\{B(x,r_x):x\in S\}$ is an open cover of $S$. Since $S$ is compact, this cover has a finite subcover; that is, there is a finite set $\{x_1,\dots,x_p\}\subseteq S$ such that $\{B(x_k,r_{x_k}):k=1,\dots,p\}$ covers $S$, i.e., $S\subseteq\bigcup_{k=0}^pB(x_k,r_{x_k})$. Now $\{r_1,\dots,r_p\}$ is a finite set of positive real numbers, so $\min\{r_1,\dots,r_p\}>0$. Let $r=\min\{r_1,\dots,r_p\}$; then $B(y,r)\cap S\subseteq B(y,r)\cap\bigcup_{k=1}^pB(x_k,r_{x_k})=\varnothing$, because $B(y,r)\cap B(x_k,r_{x_k})=\varnothing$ for $k=1,\dots,p$. To see this, suppose that $z\in B(y,r)\cap B(x_k,r_{x_k})$; then $$\|x_k-y\|\le\|y-z\|+\|z-x_k\|<r+r_{x_k}\le 2r_{x_k}=\|x_k-y\|\;,$$ which is absurd. (The step $r+r_{x_k}\le 2r_{x_k}$ is justified by the fact that $r\le r_k$.)
But then $B(y,r)$ is an open neighborhood of $y$ disjoint from $S$, contradicting the hypothesis that $y$ is an accumulation point of $S$. This contradiction shows that every accumulation point of $S$ must in fact belong to $S$ and hence that $S$ is closed. $\dashv$

Note that this is essentially the same argument that copper.hat gives in his answer; it’s just arranged to stay a bit closer to the one that you summarized incorrectly.
The intuitive idea is really fairly simple. Since $y\notin S$, we can cover $S$ with open balls so small that $y$ isn’t even in the closure of any of them; that’s what we’re doing when we choose the radii $r_x$ to be less than the distance from $x$ to $y$. (We actually took $r_x$ to be half the distance from $x$ to $y$, but that was just a convenience; anything less than $\|x-y\|$ would work.) Then we use the compactness of $S$ to say that we don’t actually need all of these open balls $B(x,r_x)$: finitely many of them are already enough to cover $S$. Let these finitely many open balls be $B(x_k,r_{x_k})$ for $k=1,\dots,p$. For $k=1,\dots,p$ let $\overline{B}(x_k,r_{x_k})=\{z\in\Bbb R^n:\|z-x_k\|\le r_{x_k}\}$, the closed ball of radius $r_{x_k}$ centred at $x_k$. We chose each $r_x$ to be less than the distance from $x$ to $y$, so $y$ is not in any of these closed balls, and therefore it’s not in their union:
$$y\notin\bigcup_{k=1}^p\overline{B}(x_k,r_{x_k})\;.\tag{1}$$
The righthand side of $(1)$ is the union of finitely many closed sets, so it is itself a closed set, and it contains $S$. Every closed set that contains $S$ also contains the closure of $S$, so $y$ is not in the closure of $S$, contradicting the hypothesis that $y$ is an accumulation point of $S$. (Or, if you arrange the argument as copper.hat did, showing directly that the complement of $S$ is open and hence that $S$ is closed.)

Finally, as copper.hat says, ‘$S$ is compact’ does not mean that $S$ has a finite open cover. It means that no matter what open cover of $S$ you start with, that open cover has a finite subcover. If you can find just one open cover of $S$ that has no finite subcover, then $S$ is not compact. Finding one open cover of $S$ that does have a finite subcover, on the other hand, tells you absolutely nothing about whether $S$ is compact.
A: $S$ is covered by $\bigcup_{x\in S}B(x,r_x)$ and as $S$ is compact, $S\subseteq  \bigcup_{k=1}^p B(x_k,r_{x_k})$. If we choose $r=\min\left\{r_{x_k}\right\}>0$ then it is true that $B(x_k,r)\cap B(y,r)=\emptyset$ and so $$S\cap  B(y,r)\subseteq \bigcup_{k=1}^p B(x_k,r_{x_k})\cap  B(y,r)=\emptyset$$ which is a contradiction as $y$ is a limit point of $S$. 
I sincerely do not understand your questions. Would you please elaborate?
A: I think it is easier and more instructive to show a) $\Rightarrow$ b) directly.
The key is that every open cover has a finite subcover. $(1,3)$ has the open cover $\{(1+\frac{1}{n},3)\}_n$, but it has no finite subcover, hence is not compact.
If $S$ is compact, then every open cover has a finite subcover. So take $\{B(0,n)\}_n$. This is an open cover of $S$, hence has a finite subcover, hence $S \subset B(0,n')$ for some $n'$. Hence $S$ is bounded.
Now suppose $y \notin S$. For each $x \in C$, there exists some $\epsilon_x>0$ such that $|y-x| \geq \epsilon_x$, or in other words, $y \notin B(x,\epsilon_x)$. Now consider the open cover $\{B(x,\frac{1}{2} \epsilon_x)\}_{x \in S}$. Since $S$ is compact, there is a finite subcover $\{B(x_i, \frac{1}{2} \epsilon_{x_i})\}_i$, and so $C \subset \cup_i B(x_i, \frac{1}{2} \epsilon_{x_i})$. By construction, if $x \in B(x_i, \frac{1}{2} \epsilon_{x_i})$, then $|y-x| \geq \frac{1}{2} \epsilon_{x_i}$, hence if we take $\delta = \frac{1}{2}\min_i \epsilon_{x_i}$, we have $B(y,\delta) \cap S = \emptyset$, and hence we have that $S^C$ is open. Hence $S$ is closed.
