Past exam of the options I course of the University of Oxford / 2009 / E / E2 / a

A subset $A$ is compact when every open cover of $A$ has a finite subcover. This may be stated using sets open in the ambient space or using the subspace topology on $A$; the definitions agree because every subspace-open set is the intersection of $A$ with an ambient-open set.

Let $(U_i{)}_{i\in I}$ be an open cover of $f(X)$, with the sets open in $Y$. The preimages $f^{-1}(U_i)$ form an open cover of $X$, because $f$ is continuous. By compactness, finitely many of these preimages cover $X$. Their corresponding sets $U_i$ cover $f(X)$, providing a finite subcover. Hence the continuous image is compact.