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This question has been answered already, this is an attempt to rephrase it.

Let $(X,\tau)$ be a cocountable topology with $X$ uncountable.

Show that $(X, \tau)$ is not Hausdorff.

Let $r\not=s$, and $r\in U$ and $s \in V$, where $U, V$ are open and not empty.

Assume $U \cap V = \emptyset$.

$X\setminus (U \cap V) =$

$(X\setminus U)\cup (X\setminus V)$.

RHS is a union of countable sets, hence countable.

LHS is $X\setminus\emptyset=X$ uncountable, a contradiction.

Perhaps nitpicking :

$\emptyset$ is open and an element of the topological space, but $X\setminus\emptyset$ is not countable.

How do you go about this?

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The complement of any non-empty open set is countable. Since $r \in U$ and $ s\in V$ your open sets $U$ and $V$ are non-empty. So your argument works fine.

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  • $\begingroup$ Thanks for pointing out explicitly that one considers $X$ \ $U$, for $U$ non-empty. $\endgroup$
    – Peter Szilas
    Commented Nov 28, 2021 at 12:12

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