[proofplan] The proof chains three facts: (1) closed subsets of compact spaces are compact ([theorem 307](/theorems/307), part 1), (2) continuous images of compact spaces are compact ([theorem 305](/theorems/305)), and (3) compact subsets of Hausdorff spaces are closed ([theorem 307](/theorems/307), part 2). Applying these in [sequence](/page/Sequence) to a [closed set](/page/Closed%20Set) $C \subseteq X$ shows $f(C)$ is closed in $Y$. [/proofplan] [step:Show $C$ is compact as a closed subset of the compact space $X$] Let $C \subseteq X$ be closed. Since $X$ is compact and $C$ is closed in $X$, the [compact subspaces and Hausdorff spaces theorem](/theorems/307) (part 1) gives that $C$ is compact. [/step] [step:Show $f(C)$ is compact as the continuous image of a compact space] Since $f: X \to Y$ is [continuous](/page/Continuity) and $C$ is compact, the [continuous image of a compact space is compact](/theorems/305). Therefore $f(C)$ is compact in $Y$. [/step] [step:Show $f(C)$ is closed as a compact subset of the Hausdorff space $Y$] Since $Y$ is Hausdorff and $f(C)$ is a compact subset of $Y$, the [compact subspaces and Hausdorff spaces theorem](/theorems/307) (part 2) gives that $f(C)$ is closed in $Y$. [guided] The three steps form a chain: \begin{align*} C \text{ closed in } X \;\xrightarrow{\text{(1)}}\; C \text{ compact} \;\xrightarrow{\text{(2)}}\; f(C) \text{ compact} \;\xrightarrow{\text{(3)}}\; f(C) \text{ closed in } Y. \end{align*} Step (1) uses compactness of $X$ and closedness of $C$. Step (2) uses continuity of $f$ and compactness of $C$. Step (3) uses the Hausdorff property of $Y$ and compactness of $f(C)$. Each hypothesis of the theorem is used exactly once: compactness of $X$ in step (1), continuity of $f$ in step (2), and the Hausdorff property of $Y$ in step (3). Removing any one hypothesis makes the conclusion false: a continuous [function](/page/Function) from a non-compact space to a Hausdorff space need not be closed (e.g., $x \mapsto 1/x$ on $(0, \infty)$ maps the closed set $[1, \infty)$ to $(0, 1]$, which is not closed). A continuous function from a compact space to a non-Hausdorff space need not be closed either. [/guided] [/step]