Halting on every input is not recognizable

Given an input $z=⟨<mi\; mathvariant="normal">⌜</mi>M<mi\; mathvariant="normal">⌝</mi>,\widehat{w}⟩$ for the complement of $HAL{T}_{TM}$ from [L4], effectively build a machine $T_{M,w}$ with input alphabet ${\mathrm{\Sigma }}_1=\{1\}$ as follows. On input $x$, the machine simulates $M$ on $w$ for exactly $\mathrm{\mid }x\mathrm{\mid }$ steps. If the simulation has not halted by then, $T_{M,w}$ halts and accepts. If the simulation has halted within those $\mathrm{\mid }x\mathrm{\mid }$ steps, then $T_{M,w}$ enters an infinite loop. This construction is effective by [L2].

If $M$ never halts on $w$, then every finite step-bounded simulation in step 1.1 remains nonhalting, so $T_{M,w}$ halts on every input $x$. Conversely, if $M$ halts on $w$ after exactly $t$ steps, then for every input $x$ with $\mathrm{\mid }x\mathrm{\mid }\ge t$ the machine $T_{M,w}$ detects that halting event during its first $\mathrm{\mid }x\mathrm{\mid }$ simulated steps and therefore loops forever. Hence $⟨<mi\; mathvariant="normal">⌜</mi>M<mi\; mathvariant="normal">⌝</mi>,\widehat{w}⟩\mathrm{\notin }HAL{T}_{TM}⟺<mi\; mathvariant="normal">⌜</mi>T_{M,w}<mi\; mathvariant="normal">⌝</mi>\in TO{T}_{TM}$.

Suppose for contradiction that $TO{T}_{TM}$ were recognizable. Then, on input $z=⟨<mi\; mathvariant="normal">⌜</mi>M<mi\; mathvariant="normal">⌝</mi>,\widehat{w}⟩$, one could compute $<mi\; mathvariant="normal">⌜</mi>T_{M,w}<mi\; mathvariant="normal">⌝</mi>$ and run the assumed recognizer for $TO{T}_{TM}$ on that code. By step 2.1, this would recognize the complement of $HAL{T}_{TM}$, contradicting [L1]. Therefore $TO{T}_{TM}$ is not recognizable, by [L3].