Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / E / 2 / a

A subset is compact if every family of open sets covering it has a finite subfamily still covering it. Equivalently, its subspace topology has the finite-subcover property.

Let $\{U_i\}$ be an open cover of $f(X)$. By continuity the inverse images $f^{-1}(U_i)$ are open and cover $X$. Compactness provides finitely many of them covering $X$, and the corresponding $U_i$ then cover $f(X)$. Thus a continuous image of a compact space is compact.