A regular normal-sum diagram has a reduced connected pair

Write $\mathrm{\Gamma }$ for the finite set of double curves. For each subset $A\subseteq \mathrm{\Gamma }$, perform just its chosen switches and take the normalization of the resulting diagram. The empty subset gives at least two components because it is parametrized by the nonempty disjoint union $X\bigsqcup Y$. The full subset gives the connected surface $F$. Choose a subset $A$ of maximal cardinality among those giving disconnected normalization. The finite list of subsets makes this a finite choice. Then $A\mathrm{\ne }\mathrm{\Gamma }$, and every strict superset of $A$ gives connected normalization.

The normalization after switching $A$ has exactly two components. Otherwise choose any remaining double curve. By [F2], its switch affects at most two components; at least one component remains unaffected and at least one affected component remains nonempty. The normalization after that additional switch would still be disconnected, contrary to maximality in step 1.1. It already had at least two components by the choice of $A$, so it has exactly two. Denote them by $G,H$.

No remaining double curve has both branches in the same one of $G,H$. If it did, its switch would leave the other component unchanged and nonempty, and the affected component would also remain nonempty by [F2]. This would again contradict step 1.1. Thus every remaining crossing is between $G$ and $H$. There are no self-intersections of either component, because any such self-intersection would lie on a remaining double curve. There are no other singularities by the diagram hypothesis. Consequently each component is an embedded normal surface: its tetrahedral disk pieces are normal by [F1], their sides match as in the partial diagram, and their interiors have no self-intersections. They are connected by their definition as normalization components.

Let $B$ be any nonempty subset of the remaining curves. Switching it gives the normalization for $A\cup B$, which is connected by step 1.1. For the separation consequence, suppose a remaining circle or arc separates both $G$ and $H$. Cutting along its preimages gives exactly two pieces of each connected surface. The crosswise switch pairs the two cut copies from $G$ with the two cut copies from $H$, producing exactly two nonempty parametrizing components. This applies to annular cutting collars of separating circles and to rectangular cutting collars of separating proper arcs. Other unswitched double curves are merely intersections of the images and do not join normalization components. Thus that one switch would have disconnected normalization, contrary to the first assertion of this step.

The partial diagram and its full resolution have the same total number of normal disks of every type by [F1]. The two normalization components therefore satisfy $v(G)+v(H)=v(F)$. All switches miss the triangulation edges, giving $w(G)+w(H)=w(F)$ as well. Each nonempty normal surface contains a normal triangle or quadrilateral and hence meets an edge, so both summand weights are strictly positive. Their sum is $w(F)$, making both strictly smaller than $w(F)$. All operations took place in the supplied neighborhoods in the original $T$, proving every assertion.