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  "body": "\n## Proof\n\nWe translate covers in each direction with the definition of the subspace topology.\n\n1. Let $X$ be a [[topological space|0CNQXVBHCX]] and let $K \\subseteq X$. Give $K$ the [[subspace topology|0D5YHQGDJY]], whose open sets are the sets $K \\cap U$ with $U$ [[open|01ZQQDCERP]] in $X$.\n2. By definition, $K$ is [[compact|00R0X1WGPX]] when $K$ with this topology is a [[compact space|0QPHW703PV]]. That is, each [[open cover|0NVPNC12JY]] of $K$ by open sets of $K$ has a finite subcover.\n3. Assume $K$ is compact. Let $(U_i)_{i \\in I}$ be open sets of $X$ with $K \\subseteq \\bigcup_{i \\in I} U_i$.\n4. For each $i$, the set $V_i = K \\cap U_i$ is open in $K$. Each $x \\in K$ lies in some $U_i$, hence in $V_i$. So $(V_i)_{i \\in I}$ is an open cover of the space $K$.\n5. By compactness, there are $i_1, \\ldots, i_n \\in I$ with $K = V_{i_1} \\cup \\cdots \\cup V_{i_n}$.\n6. Since $V_{i_j} \\subseteq U_{i_j}$, we get $K \\subseteq U_{i_1} \\cup \\cdots \\cup U_{i_n}$. This proves the first implication.\n7. Now assume the cover condition for open sets of $X$. Let $(V_i)_{i \\in I}$ be an open cover of the space $K$ by open sets of $K$.\n8. For each $i$, choose an open set $U_i$ of $X$ with $V_i = K \\cap U_i$. This is possible by step 1.\n9. Then $K = \\bigcup_i V_i \\subseteq \\bigcup_i U_i$. By the assumption, there are $i_1, \\ldots, i_n \\in I$ with $K \\subseteq U_{i_1} \\cup \\cdots \\cup U_{i_n}$.\n10. Let $x \\in K$. Then $x \\in U_{i_j}$ for some $j$, and $x \\in K$, so $x \\in K \\cap U_{i_j} = V_{i_j}$.\n11. So $V_{i_1}, \\ldots, V_{i_n}$ is a finite subcover of $(V_i)_{i \\in I}$. Thus $K$ is a compact space, and $K$ is compact.\n\n## Idea\n\nAn open set of the subspace $K$ is just an open set of $X$ cut down to $K$.\nSo a cover of $K$ by open sets of $X$ and a cover by open sets of $K$ carry the same information, and finite subcovers match up.\n",
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