An affine open in a smooth integral variety has Cartier boundary

Let $Z$ be an irreducible component of the closed complement and $\eta$ its generic point. In $\mathrm{Spec}R$, with $R={\mathcal{O}}_{X,\eta }$, the inverse image of $X\setminus U$ is just the closed point: no different component of the complement contains $\eta$. Thus the inverse image of $U$ is the punctured spectrum. The open immersion $U\to X$ is affine, because for every affine open $T\subset X$ the intersection $U\cap T$ is the pullback of the closed diagonal into $U{\times }_{k}T$, hence affine. Affineness is preserved by base change by its spectrum description. Hence the punctured spectrum of $R$ is affine.

The ring $R$ is a local UFD by [F1]. If $\dim R\ge 2$, every height-one prime remains in the punctured spectrum. A section there is an element of the fraction field regular at all such primes. In a UFD, write a fraction in relatively prime numerator and denominator; a nonunit denominator has an irreducible prime factor and yields a pole in its height-one localization. Therefore the fraction must be in $R$. Conversely every element of $R$ restricts to a section, so the punctured spectrum has global-section ring $R$. Since it is affine, [F2] identifies it with $\mathrm{Spec}R$ via its canonical restriction map, contradicting omission of the closed point. Thus $\dim R=1$; dimension zero is excluded because $U$ is dense.

The complement has finitely many irreducible components by Noetherianity, each of codimension one by step 2.1. Their sum, each with coefficient one, is an effective Weil divisor, and [F1] makes it an effective Cartier divisor with exactly the required support. If the complement is empty, take the zero divisor.