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  "body": "\n## Proof\n\nWe show that $C(x)$ is connected, then that its closure is connected, so that the closure is not larger than $C(x)$.\n\n1. Recall that $S \\subseteq X$ is [[connected|0R37DFJKFW]] when it is a [[connected space|0D939CMME7]] in the [[subspace topology|0D5YHQGDJY]]. The open sets of $S$ are the sets $S \\cap U$ with $U$ open in $X$.\n2. A set $S$ is not connected exactly when there are open $U, V \\subseteq X$ with $S \\cap U$ and $S \\cap V$ nonempty, $S \\cap U \\cap V = \\varnothing$ and $S \\subseteq U \\cup V$.\n3. The set $\\{x\\}$ is connected, since it has no two nonempty disjoint subsets. So $x \\in C(x)$.\n4. $C(x)$ is connected. Suppose not, and take $U, V$ as in step 2 for $S = C(x)$. As $x \\in C(x) \\subseteq U \\cup V$, we may assume $x \\in U$, after swapping $U$ and $V$.\n5. Let $S$ be a connected subset of $X$ with $x \\in S$. The sets $S \\cap U$ and $S \\cap V$ are open in $S$, disjoint, and cover $S$, because $S \\subseteq C(x)$.\n6. As $S$ is connected, one of them is empty. It is not $S \\cap U$, which contains $x$. So $S \\cap V = \\varnothing$, and $S \\subseteq U$.\n7. $C(x)$ is the [[union|05GK32H72H]] of all such $S$, so $C(x) \\subseteq U$ and $C(x) \\cap V \\subseteq C(x) \\cap U \\cap V = \\varnothing$. This contradicts step 2. So $C(x)$ is connected.\n8. Let $K = \\overline{C(x)}$ be the [[closure|0DRKT1QJR7]]. Then $K$ is closed and $C(x) \\subseteq K$, by [[this result|0SHC6Z304Z]].\n9. $K$ is connected. Suppose not, and take $U, V$ as in step 2 for $S = K$.\n10. Then $C(x) \\cap U$ and $C(x) \\cap V$ are open in $C(x)$, disjoint, and cover $C(x)$. As $C(x)$ is connected, one of them is empty. Say $C(x) \\cap V = \\varnothing$; the other case is the same with $U$ and $V$ swapped.\n11. Then $C(x) \\subseteq X \\setminus V$, which is [[closed|0G5AV9Q4RR]]. By the minimality of the closure in [[this result|0SHC6Z304Z]], $K \\subseteq X \\setminus V$.\n12. So $K \\cap V = \\varnothing$, which contradicts step 2. Thus $K$ is connected.\n13. $K$ is connected and contains $x$, so $K$ is one of the sets in the union that defines $C(x)$. Thus $K \\subseteq C(x)$.\n14. With step 8, $C(x) = K$. So $C(x)$ is closed, and by step 7 it is connected.\n\n## Idea\n\nThe union of connected sets through one point is connected, because any splitting would have to put each of them on the side of that point. The closure of a connected set is still connected. So the closure of the largest connected set through $x$ cannot be larger than it, which means that this set is closed.\n",
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