Let $\mathbb{A}^1$ be the one dimensional $\mathbb{C}$ affine space, and equip it with the Zariski topology. I know the topology is non-Hausdorff, but I am trying to understand if my reasoning is correct.
We have that $\mathbb{C}[x]$ is a principal ideal domain so every ideal is generated by one element, implying that the algebraic sets are specifically the roots of finite degree polynomials in $\mathbb{C}$. Therefore the closed sets are finite subsets of $\mathbb{C}$, and the open sets are the compliments. Let $U$ and $V$ be open sets in $\mathbb{C}$, then there exist some some $y_1,\dots, y_n$, $z_1,\dots, z_m\in \mathbb{C}$, such that $U=\{y_1,\dots, y_n\}^c$ and $V=\{z_1,\dots, z_m\}$. If $m=n$ and $z_i=y_i$ for all $i$ (or some permutation) then clearly these are not disjoint. If $y_i=z_i$ for some $i$, then the sets can't be disjoint, because if we take a small enough open (in the usual topology of $\mathbb{C}$) ball of $z_i$, then the set $B_{z_i}\smallsetminus\{z_i\}$ lies in both $U$ and $V$ so they are not disjoint. Now finally if $z_i\neq y_i$ for all $i$, then $z_i\in U$, in particular a small enough open ball (again usual topology) containing $z_i$ lies in $U$, so $B_{z_i}\smallsetminus z_i$ also is contained in $U$. However, $B_{z_i}\smallsetminus z_i$ also lies in $V$, hence $U$ and $V$ are not disjoint. This implies that there are no disjoint open sets in the $\mathbb{A}$ with the Zariski topology, so Hausdorffness is absolutely hopeless.
Is this reasoning correct? Is there a more direct way of showing that the Zariski topology is not Hausdorff in this case? My apologies if this question is naive, I am just starting to learn algebraic geometry.