Let $T=\{ U\subseteq X:X\setminus U\textrm{ is countable}\}\cup \{\emptyset\}$
Then this is known as co-countable topology.
Clearly,real line with co-countable topology is not Hausdorff.
For: If T is Hausdorff there exist two disjoint open sets that seperate the any pair of points;say G and H.
Then $X\setminus (A \cap B)=X$.
So the LHS is countable, but RHS is uncountable, which is a contradiction.
Can I use the same argument to prove the result if X is the complex plane or the Euclidean space?
Best Answer
All that matters for topological properties of the co-countable topology is the cardinality of the underlying set. (It is very easy to show that if $X$ and $Y$ are co-countable topological spaces and $|X| = |Y|$, then $X \cong Y$.)
So, to answer your question: yes, you can use the same argument for these sets, since they all have the same cardinality as $\mathbb{R}$.
But, even more, if $X$ is any uncountable set, then your argument shows that the co-countable topology on $X$ is not Hausdorff (you never used anything particular about the cardinality of $\mathbb{R}$ in your argument, except that it is uncountable).