Theorem
A closed subset
of a countably compact space
is countably compact.
Proof
Let
be a countably compact set and let
be a closed subset of![]()
Let
be an infinite subset of
A is also an infinite subset of a countably compact space
An accumulation point
of
exists such that![]()

Since![]()
is also an accumulation point of
Since
is closed, it contains all it's accumulation points, so
Hence an infinite subset
of a closed subset
has an accumulation point
hence
is countably compact.