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  "body": "\n## Proof\n\nWe check the open cover condition for $f(K)$ directly.\n\n1. Let $(V_i)_{i \\in I}$ be a family of open subsets of $Y$ with $f(K) \\subseteq \\bigcup_{i \\in I} V_i$.\n2. By [[compactness via open covers|0P6PMJB4MA]], it is enough to find finitely many indices $i_1, \\ldots, i_n$ with $f(K) \\subseteq V_{i_1} \\cup \\cdots \\cup V_{i_n}$.\n3. For each $i \\in I$, let $U_i = f^{-1}(V_i)$ be the [[preimage|0FGCFY1AZX]]. Since $f$ is [[continuous|0SY9ECQ19Q]], each $U_i$ is open in $X$.\n4. Let $x \\in K$. Then $f(x)$ is in the [[image|0Q3F20XB0A]] $f(K)$, so $f(x) \\in V_i$ for some $i$ by step 1. Thus $x \\in U_i$.\n5. So $(U_i)_{i \\in I}$ is a family of open subsets of $X$ whose union contains $K$.\n6. The set $K$ is [[compact|00R0X1WGPX]]. By [[compactness via open covers|0P6PMJB4MA]], there are indices $i_1, \\ldots, i_n$ with $K \\subseteq U_{i_1} \\cup \\cdots \\cup U_{i_n}$.\n7. Let $y \\in f(K)$, and write $y = f(x)$ with $x \\in K$. By step 6, $x \\in U_{i_k}$ for some $k$, so $y = f(x) \\in V_{i_k}$.\n8. Hence $f(K) \\subseteq V_{i_1} \\cup \\cdots \\cup V_{i_n}$. By step 2, $f(K)$ is compact in $Y$.\n\n## Idea\n\nPull the cover of $f(K)$ back along $f$. Continuity makes the pulled back sets open, and they cover $K$. A finite part of them covers $K$, and the matching sets cover $f(K)$.\n",
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