Regarding the Proof of the problem mentioned in the title, I followed the Proof included in this post: When is the Zariski topology $T_2$?
We know that $\mathbb{R}$ with the Zariski topology is not T2.
If $X=\mathbb{R}^n$, then we can reduce to $n=1$ considering the subspace $Y= \mathbb{R} \times \left \{0 \right \}^{n-1} \subseteq \mathbb{R}^{n}=X$. Then, the Zariski topology on $Y$ coincides with the Zariski tolopogy on $\mathbb{R}$. Let $p \in \mathbb{R}[x_1, \dots, x_n]$, the subset on which it vanishes is the subset of $\mathbb{R}$ on which the single-variable polynomial $p(x_1,0, \dots, 0)$ vanishes.
Does it work?
Best Answer
No. You showed that $Y$ is closed, which is irrelevant. You have to show that $Y$ is not Hausdorff. And the Zariski topology on $Y$ is just the cofinite topology (closed sets are only $Y$ and the finite subsets of $Y$; open sets are the complements of those.), and any two non-empty open subsets $O_1,O_2$ of $Y$, are of the form $O_1 = Y\setminus F_1$ and $O_2 = Y\setminus F_2$ where $F_1, F_2$ are finite subsets of $Y$. And as $Y \simeq \Bbb R$, $Y$ is uncountable and we have many $p \in Y$, with $p \notin F_1\cup F_2$. This $p \in O_1 \cap O_2$.
So in $Y$ all non-empty open sets intersect non-emptily, so $Y$ is even anti-Hausdorff.
Note that this immediately implies $\Bbb R^n$ is not Hausdorff either (as that property is hereditary).