A long blockade has a wide support-invariant, support-uniform minor

List the finitely many ordered trees on at most $\tau$ vertices. By repeated application of [L1], one tree at a time, choose $K$ large enough that every $K$-block blockade has a $k$-block sub-blockade whose trace is monochromatic for every tree on this list. This is a backward iteration of finitely many finite Ramsey numbers; when an order has fewer than $s$ blocks, its $s$-vertex trace is empty.

For this $K$, there are at most ${\tau }^{\tau }$ ordered-tree types on at most $\tau$ vertices: root each tree at its first ordered vertex; for size $s$ the parent map on the other $s-1$ ordered positions has at most $s^{s-1}$ possibilities, and ${\sum }_{s=1}^{\tau }{s}^{s-1}\le {\tau }^{\tau }$. Each trace has at most $2^K$ supports. Thus the sum $c_{\tau }$ of all trace cardinalities is an integer in $[0,M]$, where $M=2^K{\tau }^{\tau }$.

Start by trimming all blocks to width $W$. Choose the largest integer $t\in [0,M]$ for which an equicardinal contraction $\mathcal{C}$ of the original $K$-block blockade has width at least ${\kappa }^{t}W$ and $c_{\tau }(\mathcal{C})\le M-t$. Such a $t$ exists because $t=0$ works. If $\mathcal{C}$ were not $(\kappa ,\tau )$-support-invariant, some contraction of it of width at least $\kappa \mathrm{width}(\mathcal{C})$ would remove a support from the trace of an ordered tree of size at most $\tau$. Trim its blocks to equal size. Trimming can only remove further supports, so the new equicardinal contraction has width at least ${\kappa }^{t+1}W$ and cost at most $M-t-1$. This contradicts maximality of $t$. Hence $\mathcal{C}$ is invariant and has width at least ${\kappa }^{M}W$.

Apply the Ramsey choice of step 1.1 to $\mathcal{C}$: for each ordered tree $J$ of size $s\le \tau$, color every $s$-subset of block indices by whether it belongs to its trace, and successively retain a monochromatic sub-blockade. Keep exactly $k$ blocks. Each trace is empty or contains every $s$-subset, so this minor is $\tau$-support-uniform.

Support-invariance survives taking a sub-blockade: extend any contraction of the sub-blockade to the discarded blocks unchanged; since all original blocks of $\mathcal{C}$ have the same size, the extended contraction retains at least the same $\kappa$ fraction of its width. A copy on retained indices survives in the extended contraction if and only if it survives in the restricted one. Thus the sub-blockade is also $(\kappa ,\tau )$-support-invariant, equicardinal, and retains the width lower bound from step 2.1.