if every subset of $X$ is compact then X is finite

Question

let be $X\subset\mathbb{R}^n$ such every $A\subset X$ is compact (closed and bounded). Is true that $X$ must be finite?

I'm trying to get a counterexample because I think that this is false.

Answer 1

This holds in an arbitrary hausdorff space $X$:

If $X$ is compact, every infinite subset has a limit point.

Suppose $X$ is not finite. Then there exists an infinite subset $A$. By what is stated above, you can choose a limit point $x$ of $A$. Take it away from $A$ if it is in $A$, and denote the new set by $\tilde{A}$. It is still infinite. By assumption, $\tilde{A}$ is compact. Since $X$ is hausdorff, $\tilde{A}$ must be closed. But there exists a limit point of $\tilde{A}$ that is not in $\tilde{A}$ by construction. This is a contradiction.

OBS: Note that this does not hold if $X$ is not hausdorff. For instance, take $\mathbb{N}$ with the $\{\emptyset, \mathbb{N}\}$ topology. Every subset is compact, but $\mathbb{N}$ is infinite.