I am referring to Rudin's definition 2.32 of compactness here: A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover.
Obviously X is a subset of X itself. We also knew that any metric space is both open and closed relative to itself. Then {X} is an open cover of X that contains a finite subcover which is also {X}. It seems to me that {X} is the only open cover X has.
If it really is the case that X is compact, it would follow that X is closed and bounded. Closed indeed, but being bounded would be troublesome.
I am sure one of the above statement is false, but I don't know which one.
The requirement is that any open cover have a finite sub cover. Consider $\Bbb R.$ The set of intervals $(n,n+2)$ for $n$ an integer covers $\Bbb R$ but there is no finite sub cover.