I am considering the Zariski topology over $Q[x_1,...,x_n$] as a topology over $\Bbb R^n$, so that a set is Zariski-closed if there exists a set of polynomials $I$ such that $V= \{r \in \Bbb R^n | f(r)=0 $ for all $f \in I \}$
First off a topological space is $T_0$ if for distinct $x,y$ we have an open set that contains one and not the other, which I believe is equivalent to having a closed set that does not contain one and not the other.
Next I consider three cases 1) If $x,y$ have rational components, then I just take a polynomial $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
2) If one of $x,y$ has all rational components and the other isn't , then again I take $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
3) If both $x,y$ have irrational components, I am stuck. I would need to find a polynomial that is zero for $x$ but not $y$ (or vice versa) which is tripping me up. Any hints appreciated.
Edited to reflect that this is over $\Bbb R_n$
This is not true. In fact, if $(a_1,\cdots,a_n)$ is a point such that its coordinates are algebraically independent over $\Bbb Q$, then by definition the only closed set that contains it is the whole $\Bbb R^n$. So, any two such points are topologically indistinguishable.