Prove that the Zariski topology is not first countable on $\mathbb{R}$.
All I'm able to show right now is that all the one-point sets $\left(\{a\}\subset \mathbb{R}^n\right)$ are closed as every point $a \in $ is the zero of the polynomial $f(x) = x-a \in \mathbb{R}[x_1, \ldots, x_n]$. But how to prove that none there is a point without countable neighbourhood basis?
Let $\{U_n\}_{n\in\Bbb N}$ be a countable neighbourhood basis for $a\in\Bbb R$. Then each $U_n$ is the complement of a finite set and $\bigcap_n U_n$ is the complement of a countable set, hence is much larger than $\{a\}$ . So pick $b\in \bigcap_n U_n\setminus\{0\}$ and see that the Zariski open set $\Bbb R\setminus\{b\}$ contains none of the $U_n$ as subset.