My teacher used this example to describe separability $ \mathbb{R} $.
If there is a set $ A \subset \mathbb{R} $ and:
(i) A is dense in $ \mathbb{R} $.
(ii) A is countable.
He then chose $ \mathbb{Q} $ to show that indeed $ \mathbb{R} $ is separable by his definition under the usual topology. I then found that the complement topology on $ \mathbb{R} $ is separable and I believe that is also done by choosing $ \mathbb{Q} $. Now I feel like any topology on $ \mathbb{R} $ is separable. Is this always the case or does it fail under the K-topology?
K-topology: $\tau(\frac1n)=\{A\subset \mathbb{R}\ :\ \big(\exists N\in\mathbb{N} \big)\ \, \forall \big(n\in \mathbb{N} , n\ge N\big) \ \ \frac1n\in A\}\cup \{\emptyset\} $
What you define as the $K$-topology is not the usual $\mathbb{R}_K$ $K$-topology that Munkres defines in his book. The latter is the topology generated by all Euclidean open subsets of $\mathbb{R}$ together with the $\mathbb{R} \setminus K$ where $K = \{\frac{1}{n}: n \in \mathbb{N}^+\}$. It's his example of a Hausdorff non-regular space.
The Munkres $K$-topology is separable as $\mathbb{Q}$ is still dense in it.
The topology $\tau(\frac{1}{n})$ that you define is also separable as every non-empty open set intersects $K$ (as defined before), which is a countable set.