A subset $A\subset \mathbb R^n$ is compact iff every nested sequence of relatively closed, non-empty subsets of $A$ has non-empty intersection. This is the problem. 
Let $A\subset\mathbb R^n$. I want to show that $A$ is compact if and only if every nested sequence $\{A_n\}_{n=1}^\infty$ of relatively closed, non-empty subsets of $A$ has non-empty intersection. 
I know how to prove "$\Rightarrow$" direction, using the fact that $A$ has the Bolzano-Weierstrass property, or more simply, using a theorem relating the compactness and the finite intersection property. 
But I am not sure how to prove "$\Leftarrow$". I tried the following. I assume that $A$ is not compact, so $A$ does not have the Bolzano-Weierstrass proprty. So there is an infinite subset $B$ which does not have a limit point in $A$. So there exists a collection of open sets $\{O_x\}_{x\in B}$ such that 
$$O_x\cap B=\{x\}$$
for all $x\in B$. Then one can find a countable subset $\{x_i\}_{i=1}^\infty\subset B$, and the corresponding collection of open sets $\{O_i\}_{i=1}^\infty$ with $O_i\cap B=\{x_i\}$. Further more, one constructs a sequence of relatively closed, non-empty subsets 
$$A_n=A-\cup_{i=1}^nO_i$$.
$\{A_n\}_{n=1}^\infty$ is a nested sequence. So $\cap_{n=1}^\infty A_n\ne\emptyset$. But 
$$\cap_{n=1}^\infty A_n=\cap_{n=1}^\infty(A-\cup_{i=1}^nO_i)=A-\cup_{n=1}^\infty O_n$$
which is surely non-empty, in general, since $\{x_i\}_{i=1}^\infty$ is generally a proper subset of $B$. 
I cannot find any contradiction. Please help me out! 

After searching online, I found out that it may be useful to use the property of being sequentially compact, etc.. But I am reading Fred H. Croom's Principles of Topology, which does not discuss it. Please use some different methods. 
 A: You've successfully found a countable collection of points $\{x_i\}\subset B \subset A$ such that $\{x_i\}$ has no limit points in $A$. 
Now let $C_j = \{x_i\}_{i = j}^\infty$. 


*

*Why are the $C_j$ nested? 

*Why are the $C_j$ relatively closed?

*What's the intersection $\cap C_j$? 

A: Let $F$ be a family of subsets of $X.$ We say that $F$ has the Finite Intersection Property (F.I.P.) iff (i) $F\ne \emptyset,$  and (ii) $\;\cap G\ne \emptyset$ whenever $G$ is a finite non-empty subset of $F.$ 
Theorem: A space $X$ is compact iff $\cap F\ne \emptyset$ for every family $F$ of closed sets such that $F$ has the F.I.P.
Proof: (1). If $X$ is not compact,let $C$ be an open cover of $X$ with no finite sub-cover. Let $F=\{X\backslash c: c\in C\}.$ Then  $F$ is a family of closed sets and $F$ has the F.I.P. but $\cap F=\emptyset.$
(2). If $F$ is a family of closed sets and $F$ has the F.I.P. but $\cap F$ is empty then $\{X\backslash f: f\in F\}$ is an open cover of $X$ with no finite sub-cover, so $X$ is not compact.
