= Solution
Let $\{O_i\}$ be an <open cover> of $Z$ in its <subspace topology>. Write $O_i=Z\cap U_i$ with each $U_i$ open in $X$. Since $Z$ is a <closed set>, $X\setminus Z$ is open, and $\{U_i\}\cup\{X\setminus Z\}$ is an <open cover> of the <compact space> $X$. A finite subcover restricts to a finite subcover of $Z$. Hence \b[the closed subset $Z$ is compact].
If $\{V_j\}$ is an <open cover> of $f(Z)$ in its <subspace topology>, the inverse images under the restricted <continuous map> $f|_Z$ form an <open cover> of $Z$. Take a finite subcover; the corresponding $V_j$ cover $f(Z)$. Thus \b[the <continuous> image $f(Z)$ is compact]. Neither argument assumes a <Hausdorff space> or asserts that a <compact set> must be <closed set> in an arbitrary <topological space>.
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