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Source: published version, p. 241, Proposition 2.3.

Statement

Use the construction and notation of the shrinking-fragment iteration, and put

$$W=\bigcup_{i=1}^{\gamma}W_i, \qquad \mathcal U=\bigcup_{i=1}^{\gamma}\mathcal U_i.$$

For every outcome of the iteration, at least one of the following holds:

$$W\in\langle\mathcal H\rangle \qquad\text{or}\qquad \mathcal H\subseteq\langle\mathcal U\rangle. \tag{1}$$

Full proof

Every edge of $\mathcal H_\gamma$ has integer size at most $\ell_\gamma<1$. Hence

$$\mathcal H_\gamma=\varnothing \quad\text{or}\quad \mathcal H_\gamma=\{\varnothing\}.$$

Suppose first that $\mathcal H_\gamma=\varnothing$. Fix $S_0=S\in\mathcal H$. At step $i$, if $S_{i-1}\in\mathcal G_i$, stop. Otherwise define

$$S_i=T(S_{i-1},W_i)\in\mathcal H_i.$$

Every $S_i$ is a subset of $S_{i-1}$ by the minimum-fragment property. Since no edge survives to $\mathcal H_\gamma$, this process stops at some $i\le\gamma$. At that step, $T(S_{i-1},W_i)\in\mathcal U_i$ and

$$T(S_{i-1},W_i)\subseteq S_{i-1}\subseteq S.$$

Thus $\mathcal U$ covers $S$. Since $S$ was arbitrary, it covers all of $\mathcal H$.

Suppose instead that $\mathcal H_\gamma=\{\varnothing\}$. The invariant (3) on the iteration page, applied to the terminal empty edge, gives

$$W=W\cup\varnothing\in\langle\mathcal H\rangle.$$

This proves (1).

The published proof's evolution sentence writes $S_i\in\mathcal G_i$ at the stopping step. Since $\mathcal G_i\subseteq\mathcal H_{i-1}$, the type-correct index is $S_{i-1}\in\mathcal G_i$, used above; this is an indexing correction only and leaves the stated construction unchanged.