The answer to the original question of Majialin doesn't need so subtle concepts as totally bounded. In a Hausdorff space is clear that a closed subset of a compact is again compact.
I am not sure in the non Hausdorff case, I am not used to think in this spaces, it is not the point but, is it also true?
It is not serious, but I can answer my question given the argument for the question. It follows from the very definition.
If K is compact and F is a closed subset of K, then any open covering of F can be "extended" to an open covering of K by adding . Hence it follows compactness of F, even not assuming ithe space to be Hausdorff.
Yes, that's a very good suggestion. Any compact set, in a Hausdorff space, is closed and the intersection of two closed sets is closed so the intersection of a compact set and a closed set is compact. Then use the fact that a closed subset of a compact set is closed.
And, of course, any metric space is Hausdorff.
Yes, of course. I only introduced the non Hausdorff case because of my curiosity. And I believe the argument works, more detailedly
if , closed and ,
open in , then . If is compact then there exists finite such that . Hence , i. e. is also compact. We do not need HAusdorff as I believed in my first post, then the original problem in a metric space is a particular case, in any topological space the propertry is true, always closed intersected compact is compact (if hopefully I have not missed anything!). Sorry because my previous posts were somehow confusing.