Swan chartwise integrality glues on the actual quasi-affine open

Every $W_i$ is a nonempty open of the integral $X$, so their finite intersection contains a nonempty principal open $O$ of $X$. Apply the common-core lemma to this $O$. Intersect its nonempty open $T_O$, the avoidance open, and all finitely many $G_i$. This is a nonempty open $G\subset {\mathbb{A}}^N$, and has a $k$-point. The local Swan conditions apply to the restrictions of the very same global parameter family, since its indexed list contains the controlling sections of each chart.

Choose that parameter and let $Y$ be its scheme-theoretic zero locus in $U$. Avoidance places $Y$ in $W$. Each $Y_i=Y\cap W_i$ is smooth geometrically integral of dimension one by [F2], and is nonempty because it contains the nonempty open subset $Y\cap O$. The latter is an open subset of $Y_1$, hence irreducible, and is a nonempty open, therefore dense, in every $Y_i$. Its closure in $Y$ contains every $Y_i$, and these cover $Y$. Thus $Y$ is irreducible; it is smooth and reduced because these are local and hold on this cover. The dimension and codimension assertions follow from those on every chart.

The same common-core and open-cover argument survives every extension of $k$: each chart curve is geometrically integral, and the common core remains a nonempty open in it. The extended cover is therefore still irreducible and reduced. Consequently $Y$ is geometrically integral. This proves the assertion for the actual quasi-affine $U$ without claiming that global generation alone implies irreducibility.