Zariski type topology in real algebraic geometry

Question

Consider the following Zariski-type topology on $\mathbb{R}$: the closed sets are given by zero sets of real analytic functions. If we define a similar topology on $\mathbb{R}^n$, where the closed sets are still given by countable sets $C$ where each point in $C$ is an isolated point (with respect to the normal topology on $\mathbb{R}^n$), and $C$ is closed (with respect to the normal topology on $\mathbb{R}^n$), then is it true that $\mathbb{R}$ and $\mathbb{R}^n$ are homeomorphic?