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How to prove that Co-countable space is not a Hausdorff space?

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Do you mean an uncountable set $X$ equipped with the co-countable topology? Just find two points $p,q\in X$ that don’t have disjoint open neighborhoods. You don’t have to look hard: any two distinct points $p,q\in X$ will work. The only fact that’s needed to prove this is that the union of two countable sets is countable.

Just let $p$ and $q$ be distinct points of $X$ and let $U$ and $V$ be open neighborhoods of $p$ and $q$, respectively, and use the definition of the co-countable topology to show that $U\cap V\ne\varnothing$: that immediately shows that $X$ is not Hausdorff.