In Liu's AG textbook, Example 2.3.6 says that when looking at $ \text{Spec }\mathbb Z[T]$ and open subscheme consisting of the union of the two basis opens $U:=D(p)\cup D(T)$ for some prime $p$, $$\mathcal O_X(U)\subseteq \mathcal O_X (D(p))\cap \mathcal O_X (D(T)) = \mathbb Z[T,1/T]\cap\mathbb Z[T,1/p]=\mathbb Z[T].$$
And this implies that $\mathcal O_X(U)=\mathbb Z[T]$ (I don't see why it cannot be a proper subset). And then, using the fact that $\mathcal O_X(U)=\mathbb Z[T]=\mathcal O_X(X)$, we can conclude that $U$ is not affine using the fact that for an affine scheme $Y$ and any scheme $X$, $\hom(X,Y)\to\hom(\mathcal O_Y(Y),\mathcal O_X(X))$ is bijective.
I'm assuming a contradiction is derived by supposing $U$ is indeed affine, and taking $Y=U$, $X= \text{Spec }\mathbb Z[T]$, but I don't see exactly how.
Best Answer
Functions on $U$ already contain the polynomials. To see this, note that functions on $D(p)$ and $D(T)$ are both localizations of an integral domain, and hence the canonical map from the polynomial ring is an injection.
For the second part, if $U$ were affine, by the observation above the inclusion $j: U \to Spec(\mathbb{Z}[T])$ would be an isomorphism, being dual to the isomorphism $\mathbb{Z}[T] \to \Gamma(U, \mathcal{O}_U)$. However, I claim that $j$ cannot be surjective. Indeed notice that the complement of $U$ is the intersection $V(p) \cap V(T)$, which contains the proper ideal $(p,T)$ and is therefore nonempty.