Proof that every set in $\mathbb{A}^n$ is compact in the Zariski topology.
Let $\{ U_\alpha \}$ be a collection of open sets. Let's first pick a numerable collection of them $U_1, U_2, \dots$, so we have something better to index on. We will be done if we can find a finite collection of them such that
$$\bigcup_{r=1}^N U_r = \bigcup_{r=1}^\infty U_r.$$
(See afterword in the end why I only have $r=1,...,\infty$ here and not all $\alpha$.)
Now, since $U_r^C$ is closed, there is an ideal $J_r$ such that $U_r^C = \mathbb{V}(J_r)$. Define $I_j = \sum_{r=1}^j J_r$. Then
$$I_1 \subset I_2 \subset \dots$$
is an ascending chain of ideals, so there must be $N$, such that $I_N=I_{N+1}=\dots$.
But then
$$\bigcup_{r=1}^j U_r = \bigcup_{r=1}^j \mathbb{V}(J_r)^C = \left(\bigcap_{r=1}^j \mathbb{V}(J_r) \right)^C = \mathbb{V}(I_j)^C$$
and $\mathbb{V}(I_j)^C = \mathbb{V}(I_N)^C$ when $j\geq N$ because $I_j = I_N$. So we get the finite sub cover
$$\bigcup_{r=1}^N U_r = \bigcup_{r=1}^\infty U_r$$
I'm little hazy about the picking of numerable cover above. How about if the cover is uncountable? Also, if there are any mistakes in the rest of the proof.
Afterword:
OK, I looked here and it seems one should consider the collection of finite sub covers and then do the above argument for a chain (i.e. a sequence(?)) and so we find a maximal element by Zorn's lemma (or whether you need it or not), now the maximal element is $U:=\bigcup_{\alpha} U_\alpha$, because if there is a set that contains a point that is not in a chain's union, add that set to the chain and you get a bigger union.
