Let $X$ be a space that is countably compact, which means that every countable open cover has a finite subcover. Assume that $f$ is a continious function. Then $f(A)$ is contably compact.
So, take a space that is countably compact. Denote $A := f(X)$. Take a countable open cover of A. I have to show now that it has finite subcover. But don't have any idea on how to proceed further. Does it follow from the fact that image of open sets are open if $f$ is continious. I mean since we know that $f$ is continious function then image of finite open subcover of X will be open as well in A. Or it is not enough?
Could anyone please give some hints?