The space-diagonal machine halts

Determine $n$, form $1^n$, and run the constructor to obtain $B=s(n)$. This halts and uses $O(n+s(n))=O(s(n))$ space. Thereafter use the fixed capped simulator, excluding the timeout counter for now. Its coded input and binary bound, cap bookkeeping, representation, and scratch fit in $K=C(n+B+1)$ cells for a fixed integer $C$, by the simulator's cap interface. The integer $K$ itself can be computed in $O(n+B+1)$ space.

Let $q,k,a$ be respectively the number of states, tapes, and tape symbols of this fixed capped simulator. On its fixed input, each head is in a position below $K+1$, and each tape can be represented by its first $K+1$ symbols; all later input symbols are fixed and all later work symbols are blank. Thus it has at most $q(K+1{)}^k{a}^{k(K+1)}$ configurations. Since $K\ge 1$ and $K+1\le 2^K$, fix an integer $d$ large enough that this number is strictly below $2^{dK}$ for every $K\ge 1$. All these constants belong to the fixed simulator, independently of the decoded adversary.

Simulate at most $2^{dK}$ transitions of that capped simulator, checking halting or a cap abort at every stage. A binary counter of $dK+1$ bits suffices. If a deterministic computation repeats a configuration, its future repeats and it cannot first halt later. Consequently any halting capped computation halts before this clock expires; otherwise the procedure returns its default answer. This counts transitions of the fixed simulator itself, including its scans and scratch updates, rather than transitions of the decoded machine.

The constructor halts by hypothesis, and the later clock forces termination. Its counter uses $O(K)=O(n+s(n))=O(s(n))$ cells, as do the capped simulation and retained input. Therefore the resulting diagonal procedure is total in the asserted space, preserving every non-aborted simulated halting answer for inversion.