Let$ (X; T_{\text{cocountable}})$ be an infinite set, show that it is closed under countable intersections. Also give an example to show that $\mathcal{T}_{\text{cocountable}}$ need not be closed under arbitrary intersections.
I was looking for some feedback on my proof:
$X\setminus\bigcap_{\alpha \in I} A_{\alpha}$, where $I$ is a countable set, equals to $\bigcup_{\alpha\in I}\{X\setminus A_{\alpha}\}$. The union of countably many sets is also countable and hence, $T_{\text{cocountable}}$ is closed under countable intersections. 
An example of an arbitrary intersection could be $X \cap \emptyset = \emptyset$ and $X\setminus \emptyset = X$, which is infinite.  

Your proof is fine, but the counterexample is not.
The cocountable topology is in fact the topology generated by the cocountable sets, which turns out to be the cocountable sets and the empty set.
As a counterexample, pick $A\subseteq X$ such that $A$ and $X\setminus A$ are both uncountable, and use $A$ as an index to find an intersection of cocountable sets which is in fact uncountable.
