= Solution
For $0<\kappa\leq4$, use the allowed probability-one event that the whole <SLE> trace lies in $\mathbb H$ at every positive time. At each deterministic rational $s>0$, part (b) makes the centered mapped future a fresh <SLE> with the same property. Consequently, almost surely, for every $t>0$,
$$
\tilde\gamma_t\in\mathbb H,\qquad
\gamma_{s+t}=g_s^{-1}(\tilde\gamma_t+\xi_s)\in D_s=\mathbb H\setminus K_s.
$$
This inverse relation holds through the trace's boundary-extension construction and puts positive-time future points in the open surviving domain. It therefore excludes every contact with the old filled hull, not just contacts with a fixed old point.
Intersect these probability-one events over all positive rational $s$. If $\gamma_u=\gamma_v$ for $0<u<v$, choose rational $s$ with $u<s<v$. The first occurrence belongs to $K_s$, whereas the second lies in $D_s$, a contradiction. A return to $\gamma_0=0$ is excluded by the original positive-imaginary-part assertion. This <simplicity of SLE from boundary-avoiding restarts> proves
$$
\boxed{\gamma\text{ is almost surely a simple curve for }0<\kappa\leq4.}
$$
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