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.