I am trying to prove the statement that every countably compact subset Y of a metric space (X,d) is closed.
I am aware of the fact that, for metric spaces, countable compactness is equivalent to compactness, which would do the trick, but I do not want to use this; I want to prove the statement directly from the definition of countable compactness (which says that every countable open cover has a finite subcover).
So I guess the way to go might be to suppose that Y is not closed, so X/Y not open (which implies that there is an x in X/Y for which the open $\epsilon$-ball has non-empty intersection with Y, for all $\epsilon>0$), and to use this to find a suitable countable open cover for Y such that there is no finite subcover. Yet I have not managed to find such a countable open cover...
Help would be greatly appreciated. (But let me stress again the fact that I am trying not to use all kinds of equivalences between definitions of countable compactness.)
Suppose that $Y$ is not closed. Then there is a $p\in(\operatorname{cl}Y)\setminus Y$, and there is a sequence $\langle y_n:n\in\Bbb N\rangle$ in $Y$ converging to $p$. For $n\in\Bbb N$ let $T_n=\{p\}\cup\{y_k:k\ge n\}$, and let $U_n=X\setminus T_n$; $T_n$ is closed in $X$, so $U_n$ is open in $X$, and $\{U_n:n\in\Bbb N\}$ is easily seen to be a countable open cover of $Y$ having no finite subcover.