Any Intersection of affine open subsets is affine This is an exercise of Shafarevich's Basic Algebraic Geometry 1, Chapter 1 Section 5 Exercise 9 page 66 in the third edition:

Show that any intersection of affine open subsets is affine [Hint: If $X,Y\subset \mathbb A^n$ are closed sets, the diagonal $\Delta\subset\mathbb A^{2n}$ is a closed subset and there is an isomorphism $X\cap Y\cong (X\times Y)\cap \Delta$]

I admit the first thing that's tripping me up is what the author means by "is affine". Does he mean affine closed or open?
Second, I don't see immediately how something like an infinite intersection would work here, especially considering how the hint would suggest looking at an infinite Cartesian product.
Third, I don't see what's wrong with this argument: If $U_1,U_2\subset \mathbb A^n$ are open affine subsets, then they can be written as $U_i=\mathbb A^n\backslash V_i$ with $V_i$ closed affine. Then $U_1\cap U_2=\mathbb A^n\backslash (V_1\cup V_2)$ and it is hence an affine open set.
Please avoid mention of schemes or cohomology since this book has not (and will not) introduce these concepts.
 A: This must be a poor translation from the original or just a question that does not sound like what it was supposed to. For example, it would be reasonable to say this question implies that $\mathbb{Z}=\bigcap_{\lambda\in\mathbb{C}\setminus \mathbb{Z}} (\mathbb{A}_\mathbb{C}^1\setminus\mathbb{V}(x-\lambda))$ is affine, but the only countable affine varieties over $\mathbb{C}$ are finite.
So it's probably just saying that the intersection of finitely many open affines is affine, i.e., isomorphic to a closed algebraic set in some affine space. But your argument doesn't work as is. It's not true that the complement of any closed algebraic set is affine. For example, $\mathbb{A}^2 \setminus\{(0,0)\}$ is not affine (i.e., as a quasi-affine variety, it is not isomorphic to any closed subset of any affine space).
Supplemental hint: show that the (open!) subset $U\times V\subset \mathbb{A}^n\times \mathbb{A}^n$ (the right hand side is the product as varieties, and the left hand side is just an open subset thereof, so far) with the induced structure of a quasi-affine variety is actually the product of $U$ and $V$ as varieties. Since the product of affines is affine, and $\Delta\cap (U\times V)$ is closed in $U\times V$, it too is affine, and it is also isomorphic to $U\cap V$.
