A subset $E \subset X$ of a topological space $X$ is dense if $\overline{E} = X$ where
$$ \overline{E} = \bigcap \lbrace C \subseteq X \mid C \text{ is closed and } E \subseteq C \rbrace $$
But in the cofinite topology closed sets are defined to be finite sets. So if $X$ is infinite and a subset $E$ is dense, then this would imply that $X$ (a infinite set) is the intersection of finite sets. Does this mean that $X$ endowed with the cofinite topology has no dense subsets?
There's one thing you're missing: there is exactly one infinite closed set available.
(Also no finite $C$s can contain $E$ if it's infinite so no finite sets will be used in the $\bigcap$.)
To first answer the last part of your question, $\mathbb X$ does has dense subsets. In fact every infinite subset of $\mathbb X$ is dense in $\mathbb X$. What $\mathbb X$ doesn't have are finite dense subsets (think about it!!).
To remove your confusion where you claim that if $\overline E=\mathbb X \implies \mathbb X$ is finite, because it is an arbitrary intersection of closed (hence finite) sets, I would like to point out that the only closed set that contains $\mathbb E$ which is dense (hence infinite) is $\mathbb X $ itself.
