TITLE: Relatively countably compact subsets without countably compact closure.
QUESTION [9 upvotes]: I am looking for a (Hausdorff or better) space $X$ and a subset $A$ of $X$ that is relatively countably compact (every sequence from $A$ has an accumulation point in $X$) such that the closure of $A$ is not countably compact.
It is known that in many "nice" spaces such examples do not exist (a classical case being normed spaces in their weak topology).
Edit: for $T_4$ spaces this cannot happen, as the closure of a relatively countably compact subset is pseudocompact (Suppose A is relatively countably compact.
If f from cl(A) to R is unbounded then f|A is unbounded as well, as A is dense in cl(A).
So we can find {x_n: n in N} in A such that |f(x)| >= n.
Let p in cl(A) be an accumulation point of this set, by A being relatively countably compact.
Then continuity at f implies that |f(x_n)| <= |f(p)| + 1, for all but finitely many n.
This contradicts the choice of the x_n, contradiction.)
So cl(A) is pseudocompact, and hence in a normal space, countably compact.
This explains the properties of the example given below.

REPLY [4 votes]: To fix Pietro's idea: take a Maximal Almost Disjoint family on $\omega$, i.e., a family $\mathcal{A}$ of infinite subsets such that any two distinct elements have a finite intersection and such that it is maximal with respect to this property. Topologize $\omega\cup\mathcal{A}$ by making each $n\in\omega$ isolated and have $\bigl\lbrace\lbrace A\rbrace\cup(A\setminus F): F$ finite$\rbrace$ as a local base at $A\in\mathcal{A}$.
By maximality $\omega$ is relatively countably compact; its closure is the whole space, which is not countably compact. 
(This is generally know as the/a Moore Mrowka space, a $\psi$-space, or $\Psi$-space ...)