A topological space $X$ is called quasi-compact if whenever ${(U_i)}_{i∈S}$ are a family of
open subsets such that $∪_{i∈S}(U_i) = X$ then there are a finite number of $(U_i)$’s which actually
cover X.
I'm thinking first we show that if $(U_{fi})$, $i ∈ S$ (where ${f_i}$, $i ∈ S$ a collection of elements of $R[X]$) is a family of principal open subsets which cover an affine
variety $X$, then there is a finite number which cover $X$.
Then we use the fact that principal open subsets are a basis for the Zariski topology and above statement to show
that a finite number of the $(U_j)$ are sufficient to cover $X$.
Best Answer
Every descending chain of Zariski closed subsets in $X$, $$X_1 \supseteq X_2,...$$
corresponds by inclusion reversal to $$I(X_1) \subseteq I(X_2)...$$
However, we know that Polynomial rings are Noetherian, so the ascending chain of ideals stabilizes.
Hence, every descending chain of closed sets does as well. By taking complements, we see that every ascending chain of open subsets stabilizes.
Let $\{B_i\}$ be some open covering of $X$. Consider the open sets that can be written as the finite union of elements in our open cover:$$\mathcal{A}:=\{ U \mid U=\bigcup_{i=1}^{k}B_i\}.$$
By the axiom of dependent choice, we know that if $\mathcal{A}$ does not contain $X$, then there exists some infinite ascending chain of open sets, which is a contradiction.