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.
Best Answer
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 $U\cap V$.