Does this characterize compactness? Recall that a collection of sets $\mathcal{A}$ has the finite intersection property if for all finite $\mathcal{B} \subseteq \mathcal{A}$ it holds that $\bigcap \mathcal{B} \neq \emptyset$.
In terms of the finite intersection property, we can define compactness as follows.

Definition. A topological space $X$ is compact iff any collection of closed
  subsets of $X$ with the finite intersection property has nonempty
  intersection.

Now personally, my intuition about this definition is pretty mediocre, so lets deduce a more intuitive theorem from the above definition.
First, some terminology. For any collection of sets $\mathcal{A}$, lets call $\mathcal{A}$ nested iff for all $A,B \in \mathcal{A}$ it holds that $A \subseteq B$ or $B \subseteq A$. Furthermore, lets say that $\mathcal{A}$ is ungrounded iff $\emptyset \notin \mathcal{A}$.
Then every nested ungrounded collection necessarily satisfies the finite intersection property.
Proof. Let $\mathcal{A}$ denote a nested ungrounded collection and suppose $\mathcal{B} \subseteq \mathcal{A}$ has finite cardinality. Then $\mathcal{B}$ inherits nested and ungroundedness. Thus $\mathcal{B}$ is nested and finite, so $\bigcap \mathcal{B}$ equals the least element of $\mathcal{B}$. Thus $\bigcap \mathcal{B} \in \mathcal{B}$. But since $\mathcal{B}$ is ungrounded, it follows that $\bigcap \mathcal{B} \neq \emptyset$.
Thus we obtain the following theorem.

Theorem. If a topological space $X$ is compact, then any nested ungrounded collection of closed subsets of $X$ has nonempty intersection.

My question is, does the converse of the above statement hold?
 A: Yes. The question can also be formulated with open sets and instead. And we can also simplify it by taking the contraposition. Besides, nested families are commonly known as chains.

A space $X$ is compact iff for every chain of open sets $\{U_i\}$ which covers $X$ we have $X=U_i$ for some $i$.

Proof. "$\Rightarrow$" is clear. "$\Leftarrow$": Assume that $\{U_i\}$ is some open cover of $X$ which does not have a finite subcover. We may assume that it is indexed by some limit ordinal $\alpha$. Choose $\alpha$ to be minimal. The open subsets $\cup_{i<\beta} U_i$ with $\beta<\alpha$ form a chain of open subsets which cover $X$. Thus, there is some $\beta<\alpha$ with $\cup_{i<\beta} U_i=X$. Then $\{U_i\}_{i<\beta}$ is a cover which does not have a finite subcover. This contradicts the minimality of $\alpha$. QED
More generally, let $P$ be any complete partial order. Call $x \in P$ compact iff $\sup_{i \in I} u_i=x$ implies that $\sup_{i \in F} u_i=x$ for some finite subset $F \subseteq I$. Then $x$ is compact iff for every chain $\{u_i\}$ in $P$ with $\sup_{i \in I} u_i = x$ we have $u_i = x$ for some $i$.
A: The paper A lemma on direct sets and chains by Bruns  answers this question.  (See Application 1 on the second page.)
Also see Problem H on p. 163 of Kelley's General Topology, which gives this as a (rather challenging) exercise.
Note that all of these proofs rely on the Axiom of Choice or one of its equivalents (well ordering, Zorn's lemma, maximal principle).  In fact, if my reasoning below is correct, then this theorem also implies the Axiom of Choice and therefore is equivalent to it.
The idea is to imitate this proof that Tychonoff's theorem implies the Axiom of Choice.
Using notation from that proof, that argument seems to go through if you use a slightly different topology on $X_i$, namely one where there are exactly two nonempty proper open sets, $A_i$ and $\{i\}$.  It is straightforward to verify that the product space has the property that given any chain of open sets that covers the product, one of the open sets must equal the product.   Hence the product is compact.  Now proceed as in the proof that Tychonoff's theorem implies the Axiom of Choice.
