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  "body": "\n## Proof\n\nWe define the subset by recursion, with a choice function.\n\n1. Let $A$ be an [[infinite set|01MKW4R283]]. Let $X$ be the set of nonempty subsets of $A$.\n2. By the [[axiom of choice|0HTGB13FE2]], there is a [[choice function|0YXEQWRGP0]] $c$ on $X$. So $c(S) \\in S$ for each nonempty $S \\subseteq A$.\n3. Let $F \\subseteq A$ be a [[finite set|0BBTH3YSZA]]. Then $A \\setminus F$ is nonempty. Otherwise $A = F$, and $A$ would be finite.\n4. Let $\\Phi$ be the set of finite subsets of $A$. Define $g \\colon \\N \\times \\Phi \\to \\Phi$ by $g(n, F) = F \\cup \\{c(A \\setminus F)\\}$. This set is finite, since it adds one element to $F$. So $g$ is well defined.\n5. By [[definition by recursion|0DSHAVH45F]], there is a function $n \\mapsto F_n$ from $\\N$ to $\\Phi$ with $F_0 = \\emptyset$ and $F_{n+1} = g(n, F_n)$.\n6. Define $f \\colon \\N \\to A$ by $f(n) = c(A \\setminus F_n)$. Then $f(n) \\in A \\setminus F_n$ and $F_{n+1} = F_n \\cup \\{f(n)\\}$.\n7. By induction on $n$, $F_n = \\{f(0), \\dots, f(n-1)\\}$. This holds for $n = 0$, and step 6 gives the inductive step.\n8. Let $m \u003c n$ be natural numbers. Then $f(m) \\in F_n$ by step 7. But $f(n) \\notin F_n$ by step 6. So $f(m) \\neq f(n)$.\n9. Hence $f$ is [[injective|0PVG1CRAJN]].\n10. As a function onto its image, $f \\colon \\N \\to f(\\N)$ is injective and surjective, so it is a [[bijection|0SNQEF7Z53]].\n11. Thus $f(\\N) \\subseteq A$ is [[countably infinite|0SEZ3JGZP7]], which proves the statement.\n\n## Idea\n\nPick elements of $A$ one at a time, each different from the ones already picked. The supply never runs out, because only finitely many elements are used at each stage and $A$ is infinite. A single choice function makes all the picks at once, and recursion strings them into a sequence.\n",
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