Confusion on Compact Space Definition. If for any open cover $\mathcal U$ of $X$, there exists a finite subcover $\mathcal V$ of $\mathcal U$, we call $X$ is compact.

Theorem 1. Let $X$ be compact. If $\{F_n\}_{n\in\mathbb N}$ is the decreasing nonempty closed sets in $X$, $\cap_{n\in\mathbb N}F_n\not=\emptyset$.  
The theorem is direct to prove, and if we substitute total ordered set for $\mathbb N$, we can also draw similar conclusion. My question is that after substitution, is the converse true?    
Question 1. For any total ordered set $\Lambda$, and $\{F_\lambda\}_{\lambda\in\Lambda}$ is the decreasing nonempty closed sets in $X$, $\cap_{\lambda\in\Lambda}F_n\not=\emptyset$ $\Longrightarrow$ $X$ is compact?

Theorem 2. Let $X$ be compact set, and $A$ be an infinite set in $X$. Then $A$ has a limit point.
Question 2. For any $A$ be an infinite set in $X$, $A$ has a limit point $\Longrightarrow$ $X$ is compact?

Could someone give some advice? Better with some examples. Thank you. 
 A: Question 1
You actually get a stronger version of compactness from your stated principle:

(★) Given a totally ordered set $\Lambda$ and $\{ F_\lambda \}_{\lambda \in \Lambda}$ a decreasing family of nonempty closed subsets of $X$, then $\bigcap_{\lambda \in \Lambda} F_\lambda \neq \varnothing$.

First, to see that (★) implies compactness, we prove by induction on the ordinal $\alpha > 0$ that if $\{ F_\xi : \xi < \alpha \}$ is a family of closed subsets of $X$ with the finite intersection property, then $\bigcap_{\xi < \alpha} F_\xi \neq \varnothing$.


*

*For $\alpha < \omega$ (finite ordinals), this follows because of the finite intersection property itself;

*Suppose the result is true for some $\alpha$, and let $\{ F_\xi : \xi < \alpha + 1\}$ be a family of closed sets with the f.i.p.  Then $\{ F_\alpha \cap F_\xi : \xi < \alpha \}$ is also a family of closed sets with the f.i.p., and so $\bigcap_{\xi < \alpha} ( F_\alpha \cap F_\xi ) = \bigcap_{\xi < \alpha + 1} F_\xi$ is nonempty by the induction hypothesis.

*If $\alpha > 0$ is a limit ordinal, and the result is true for all $\beta < \alpha$, then $E_\beta := \bigcap_{\xi < \beta} F_\xi$ is a nonempty closed set for all $\beta < \alpha$, and clearly $\beta < \gamma < \alpha$ implies $E_\gamma \subseteq E_\beta$.  It thus follows that $\bigcap_{\beta < \alpha} E_\beta \neq \varnothing$ (by property (★) of $X$), and it is easy to show that $\bigcap_{\beta < \alpha} E_\beta = \bigcap_{\beta < \alpha} F_\beta$.


So given a family $\mathcal{F}$ of closed subsets of $X$ with the f.i.p., if we enumerate it as $\{ F_\xi : \xi < \alpha \}$ for some ordinal $\alpha$ it follows that $\bigcap \mathcal{F} = \bigcap_{\xi < \alpha} F_\xi \neq \varnothing$.  Therefore $X$ is compact.
But, as I said above, we get a stronger property from this property.  Consider the following definitions:

Definition: By a net in a topological space $X$ we mean a function $f : \Sigma \to X$ where $\Sigma$ is any directed set.  We often denote nets by $\langle x_\sigma \rangle_{\sigma \in \Sigma}$.  
  
  
*
  
*A net $\langle x_\sigma \rangle_{\sigma \in \Sigma}$ is said to converge to $x \in X$ if for each open neighbourhood $U$ of $x$ there is a $\tau \in \Sigma$ such that $x_\sigma \in U$ for all $\sigma \in \Sigma$ with $\tau \leq \sigma$.
  
*By a subnet of a net $\langle x_\sigma \rangle_{\sigma \in \Sigma}$ we mean a net of the form $\langle x_\sigma \rangle_{\sigma \in T}$ where $T \subseteq \Sigma$; it is called cofinal if for each $\sigma \in \Sigma$ there is a $\tau \in T$ with $\sigma \leq \tau$.
  
*A net $\langle x_\sigma \rangle_{\sigma \in \Sigma}$ is called linearly ordered if $\Sigma$ is a totally ordered set.
  
  
  Definition. A topological space $X$ is called chain compact if every linearly ordered net in $X$ has a convergent cofinal subnet.

I claim that every topological space satisfying your property (★) is actually chain compact.  Suppose that $\langle x_\sigma \rangle_{\sigma \in \Sigma}$ is a linearly ordered net in $X$.  For each $\sigma \in \Sigma$ define $F_\sigma = \overline{ \{ x_\tau : \tau \geq \sigma \} }$.  Clearly $\langle F_\sigma \rangle_{\sigma \in \Sigma}$ is a descending linearly ordered family of nonempty closed subsets of $X$, and so by (★) it follows that $\bigcap_{\sigma \in \Sigma} F_\sigma \neq \varnothing$.  Taking any $x \in \bigcap_{\sigma \in \Sigma} F_\sigma$, by properties of Moore-Smith convergence it follows that there is a cofinal subnet $\langle x_\sigma \rangle_{\sigma \in T}$ which converges to $x$.
It should be noted that some very familiar compact spaces are not chain compact: one example is the closed unit interval $[0,1]$.  (If $\langle x_\xi \rangle_{\xi < \omega_1}$ is any one-to-one $\omega_1$-sequence in $[0,1]$, it follows that no cofinal subnet is convergent (there are convergent subnets, but these are not cofinal).)
Question 2
As mentioned in the comments, this is false.  The property you have given is called limit point compactness.  An in general stronger property called countable compactness is equivalent to it in the realm of T1 spaces. While every compact space is limit point compact, the converse does not hold.
The prototypical counterexample would be the ordinal space $\omega_1 = [ 0 , \omega_1 )$.  If $A \subseteq [0 , \omega_1)$ is infinite, then we can  find a strictly increasing sequence $\langle \alpha_n \rangle_{n \in \mathbb{N}}$ in $A$.  It follows that $\alpha = \sup_{n \in \mathbb{N}} \alpha_n$ is itself a countable ordinal, and not too difficult to show that it is a limit point of $A$.  However $[ 0 , \omega_1 )$ is not compact: the family of open sets $[ 0 , \alpha )$ for $\alpha < \omega_1$ comprises an open cover with no finite (actually, no countable) subcover.
