Solution
= Solution
\b[True.] Let $\{V_\alpha\}$ be an <open cover> of $f(S)$ in its <subspace topology>. Write $V_\alpha=f(S)\cap W_\alpha$ with $W_\alpha$ open in $Y$. Since $f$ is <continuous>, the sets $S\cap f^{-1}(W_\alpha)$ are open in $S$ and cover $S$. By <compactness>, finitely many cover $S$. Their corresponding $V_\alpha$ then cover $f(S)$, proving that the <continuous image of a compact space> is compact.