Question: Do you know if compactness of a set, in a metric space, can be expressed in terms of closed sets?

That is, Is it true the following statement: Let $(X,d)$ a metric space and $A\subseteq X$ such that every closed cover of A has a finite subcover. Then A is compact (in the usual sense).

I think it is false, so I'm trying to find a non compact set $A$ on a metric space $X$ such that, whenever $\{F\_\lambda\}$ is a collection of closed sets such that $A\subseteq\cup\_\lambda F\_\lambda$, there are $\lambda\_1,...,\lambda\_n$ with $A\subseteq\cup\_{i=1}^n F\_{\lambda\_i}$.

Answer: 
Since points are closed subsets of a metric space, a subset of a metric space (or, more generally, of a T1 space) has your property if and only if it is finite. Of course, finite sets are compact.

A subset $A$ of a topological space $X$ is compact in the usual sense if and only if for all families $\mathcal F$ of closed sets such that $A\cap \bigcap\mathcal F=\emptyset$ there are $F\_1,\cdots, F\_n\in \mathcal F$ such that $A\cap F\_1\cap\cdots\cap F\_n=\emptyset$.
The usual dual formulation of compactness in terms of closed sets: $X$ is compact iff for every family of closed sets $\mathcal{C}$ of $X$ such that $\bigcap \mathcal{C}=\emptyset$, there are finitely many $C\_1,\ldots C\_n$ from $\mathcal{C}$ such that $C\_1 \cap C\_2 \cap \ldots \cap C\_n =\emptyset$. This holds in all topological spaces.