 Solution

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.