Show that $\mathbb{A}^n$ on the Zariski Topology is not Hausdorff, but it is $T_1$

Since Seth has proved that $\mathbb A^n$ is $T_1$, I'll prove that $\mathbb A^n$ is not Hausdorff [Seth correctly pointed out that Jen's proof is incorrect for $n\gt1$] .

By considering complements of open subsets Hausdorffness amounts to proving that the union of two closed strict subsets $X_1\subsetneq \mathbb A^n, X_2\subsetneq \mathbb A^n$ is not the whole affine space: $X_1 \cup X_2\subsetneq \mathbb A^n$.
Now, writing $X_i=V(I_i)$, the inequality $X_i \neq \mathbb A^n$ is equivalent to $f_i\in I_i$ for some non zero $f_i\in k[T_1,\dots,T_n]$.
We have $X_1\cup X_2=V(I_1\cdot I_2)\subset V(f_1\cdot f_2)$, so that it is enough to prove $V(f_1\cdot f_2)\neq \mathbb A^n$ to obtain the desired conclusion $X_1 \cup X_2\neq \mathbb A^n$.
But $V(f_1\cdot f_2)\neq \mathbb A^n$ is clear: it results from the well-known fact that if $k$ is an infinite field a non-zero polynomial $g(T_1,\dots,T_n)\in k[T_1,\dots,T_n] $ (here $g=f_1\cdot f_2$) cannot vanish on all $(a_1,\dots,a_n)\in k^n$ .

$T_1$ is equivalent to single point sets being closed.

Let $(a_1,\dots,a_n)\in\mathbb{A}^n$. Consider the zero locus of the polynomials $x_i-a_i$ where $i$ ranges from $1$ to $n$.

For your proof of non-Hausdorffness it looks like you are assuming that $\mathbb{A}^n$ has the cofinite topology, which is not true in general. $\mathbb{A}^1$ does have the cofinite topology, but $\mathbb{A}^2$ does not, since for example, the polynomial $x\in\mathbb{k}[x,y]$ has infinitely many zeros.