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The punctured affine plane $U={\mathbb{A}}_k^2\setminus \{(0,0)\}$ is not an affine scheme. On its cover $D(x)\cup D(y)$, global sections are $k[x,y{]}_x\cap k[x,y{]}_y=k[x,y]$ inside the field of fractions. Coprimality of $x$ and $y$ proves this equality. If $U$ were affine, its inclusion into ${\mathbb{A}}_k^2$ would be an isomorphism by affine scheme reconstruction from global sections, contradicting the omitted point.

Put $A=k[t_1,t_2]$. The punctured affine plane is covered by the principal open subschemes $D(t_1)$ and $D(t_2)$. Since $A$ is an integral domain, restriction identifies its global sections withThis intersection is $A$: if $a/t_1^r=b/t_2^s$, then $t_2^{s}a=t_1^{r}b$, and unique factorization forces $t_1^r$ to divide $a$. Therefore the inclusion $j:X\hookrightarrow \mathrm{Spec}A$ induces the identity on global regular functions.

If $X$ were an affine scheme, affine scheme reconstruction from global sections would make $j$ an isomorphism of schemes, because its global-section map is an isomorphism. But its image omits the closed point $(t_1,t_2)$. The punctured affine plane is therefore not affine. This is a useful example of global regular functions failing to distinguish a nonaffine open subscheme from its affine ambient space.