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  "body": "\n## Counterexample\n\nLet $X$ be an infinite set and let $p \\in X$.\nDeclare $U \\subseteq X$ open if $p \\notin U$, or if $p \\in U$ and $X \\setminus U$ is finite.\nThis is the Fort space on $X$ with point $p$.\nIt is [[compact|0TVJZYB6SH]] and [[Hausdorff|0S1FJ8VEX9]].\nIf $X$ is [[uncountable|0W7FBYS1F1]], it is not [[first-countable|0CP5NDD21C]], and so not [[metrizable|091GJ3FYD7]].\n\n## Remarks\n\nCompactness: in an open cover, one member $U$ contains $p$, and the finitely many points outside $U$ need finitely many more members.\nHausdorff: two points $x, y \\ne p$ are separated by $\\set{x}$ and $\\set{y}$, and $x \\ne p$ is separated from $p$ by $\\set{x}$ and $X \\setminus \\set{x}$.\nLet $X$ be uncountable and let $U_1, U_2, \\dots$ be open sets containing $p$.\nTheir intersection has countable complement, so it contains a point $x \\ne p$.\nThen $X \\setminus \\set{x}$ is an open set containing $p$ that contains no $U_n$, so $p$ has no countable neighbourhood base.\nThe Fort space is the [[one-point compactification|0GWN3CMRK8]] of the discrete space $X \\setminus \\set{p}$.\nIf $X$ is countably infinite, it is homeomorphic to $\\set{0} \\cup \\set{1/n : n \\ge 1}$ with the usual topology, which is metrizable.\n\n## Explanations\n\n### L1 human\n\nTake a set of isolated points and add one extra point $p$.\nA set around $p$ is open when it leaves out only finitely many points.\nThis makes the space compact and keeps points apart.\nWith uncountably many points, no countable list of open sets can pin down $p$, so no distance function gives this topology.\n",
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