Let $X$ be an uncountable set. We define the cocountable topology $\tau$ as the set of all subsets $U\subseteq X$ such that either $U=\emptyset$ or $X\setminus U$ is countable.
I am interested in the conditions under which a set $E\subseteq X$ will be dense in $X$, and I honestly don't know where to start.
I have already shown that $(X,\tau)$ defines a topological space. I know that by definition, a set $E$ is dense in $X$ with respect to $\tau$ if for every $x\in X$ and every open set $U \subseteq X$ with $x\in U$, there exists an element $y\in E$ such that $y\in U$, or equivalently that $E$ is dense in $X$ with respect to $\tau$ if for every nonempty open set $U\subseteq X$, $E\cap U \neq \emptyset$.
I would like to see a complete proof if possible.
EDIT: I first want to show that: If $E\subseteq X$ is dense, then $X\setminus E$ is countable.
I attempted to prove this by the contrapositive (i.e. show that if $X\setminus E$ is uncountable, then $E$ is not dense in $X$). Suppose that $X\setminus E$ is uncountable. Then, $X \setminus E$ is open with respect to $\tau$, and so by definition $X\setminus (X\setminus E)=E$ is closed in $X$ with respect to $\tau$. Then, $E=\overline {E}$, where $\overline {E}$ denotes the closure of $E$. Since $E$ is countable and $X$ is uncountable, $E\neq X$. I am struggling to jump from here to the fact that $E$ is not dense in $X$.
A subset $A$ of $X$ (having the co-countable topology) is closed iff $A = X$ or $A$ is at most countable (so including finite). This is straight from the definition of the co-countable topology.
The closure $\overline{A}$ of a set $A$ is the smallest closed set $C$ such that $A \subseteq C$.
If $A$ is at most countable, it is already closed, so its closure is $A$ again (It cannot be smaller than that) and as $A \neq X$ (recall that $X$ is assumed to be uncountable), $A$ is not dense.
If $A$ is uncountable, then the only closed set of $X$ that can contain $A$ is $X$ (if $C$ is at most countable (the only other type of closed set), we cannot have $A \subseteq C$, as any subset of a set that is at most countable is again at most countable, which $A$ is not), so $\overline{A} = X$ for $A$ uncountable (there is only one candidate closed set) and thus $A$ is dense.
So the situation is atypical (for a general space): a subset is either "small" (at most countable) and closed, or "big" (uncountable) and dense.