I want to prove that the topology on $\mathbb{R}_K$ satisfies the Hausdorff axiom.
We know the topology on $\mathbb{R}_K$ is generated by basis $(a,b)$ and $(a,b)-K$ where $K =\{1/n\}_{n \in \mathbb{Z}_+}$.
My attempt:
Let $a$ and $b$ are two distinct points in $\mathbb{R}$. Then there exists two disjoint neighborhoods $U$ and $V$ which are open in the standard topology. Since the topology on $\mathbb{R}_K$ is finer than the standard topology, $\mathbb{R}_K$ is Hausdorff.
Is the proof correct?
That is perfectly fine. Indeed it holds in general: Every time you have a Hausdorff topological space $(X, \tau)$ , every time you endow $X$ with a finer topology $\tau'$, then $(X, \tau')$ is Hausdorff. The main idea is that for every $2$ distincts points you just need to find at least one neighborhood for every point that contains it and that is disjoint from the other.