Filling constants give a uniform slimness bound

First take the given countable family. Choose a basepoint in each nonempty space and form its wedge. F2 gives $m_W(P)\le F(P)$, so $m_W(P)\mathrm{/}P\to 0$. F3 applies to this geodesic wedge and yields a finite number $\delta$ bounding every triangle's slimness there. Each chosen factor triangle has precisely its original metric on the union of its sides, since the factor embeds isometrically. Distances from a side point to the union of the other two sides are therefore unchanged. The same $\delta$ works in every factor. An empty family satisfies the conclusion with $\delta =0$.

For the presentation assertion suppose no common bound exists. Finite generating alphabets can be relabelled by $\{1,\dots ,k\}$, and finite sets of finite words over these alphabets form a set. Take each presented group as the quotient of the corresponding set of words and each edge realization using copies of $[0,1]$. Thus these spaces and the collections of their interval-parameterized triangles are sets, so the following choices are legitimate set-indexed applications of AC. For each positive integer $n$, failure of a common bound supplies one such presentation and a chosen triangle whose slimness exceeds $n$; choose them, and point each space at its identity vertex. Relabelling preserves word lengths, area and distances.

All these chosen spaces have the identical function $F(P)=A(K,L)\sqrt{P+7}+B(L)$ by F1. Its nonnegativity, monotonicity and sublinearity were established there with constants depending only on the fixed $K,L$. Apply step 1.1 to this countable family. It gives a finite $\delta$ bounding the selected triangle in each factor. A positive integer $n>\delta$ contradicts the choice of a triangle with slimness greater than $n$. Therefore a common finite bound exists for the entire class at these $K,L$; denote one by $\delta (K,L)$. If no presentations satisfy the hypotheses, zero is a bound. This proves both assertions.