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I can just figure out that arbitrary uncountable set has a countable subset, which is trivial. However, I don't know whether I can get a bounded one.

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If we reject the Axiom of Choice, then it is possible that there is an infinite set of real numbers that has no countable subset. – Henning Makholm Sep 21 '12 at 11:33
up vote 14 down vote accepted

Yes, you can. You can also get a bounded uncountable subset.

Let $X$ be your uncountable subset of $\Bbb R$. For each integer $n$ let $X_n=X\cap[n,n+1)$. Then $X=\bigcup_{n\in\Bbb Z}X_n$, and a countable union of countable sets is countable, so at least one of the sets $X_n$ must be uncountable. That $X_n$ is then a bounded, uncountable subset of $X$. Now just pick a countably infinite subset of $X_n$, and you have your bounded, countably infinite subset of $X$.

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