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.  

I concur with the countable intersections, the equivalent formulation that the closed sets are closed under countable unions is even easier.
As to arbitrary intersections, your examples make no sense (the results are both in the topology?). 
Instead consider $X_x = X\setminus \{x\}$ for all $x \in X$, all of which are open. Suppose $X$ is uncountable (or else the topology is discrete and closed under all intersections) and write $X$ as a disjoint union $A \cup B$, which are both uncountable. Then $B$ is not open (why?) and $B = \cap_{x \in A} X_x$.
