= Solution
Let $R=X(l)$ and $C$ be its complement, with Hilbert-space dimension $d_C$. Use <Haar twirling conditional expectation> to define
$$
A_R(t)=\frac{\operatorname{Tr}_C A_X(t)}{d_C}\otimes I_C
=\int dU_C\,(I_R\otimes U_C)A_X(t)(I_R\otimes U_C^\dagger).
$$
This is an operator supported on $R$. The normalization $1/d_C$ is essential: the <partial trace> alone would not fix an operator already supported on $R$.
Subtract the integrand from $A_X(t)$ and use the <operator norm> triangle inequality. Since $A-U A U^\dagger=[A,U]U^\dagger$ and every <unitary operator> has norm one, the assumed <Lieb-Robinson bound>, applied to a unitary supported on the entire complement, gives
$$
\|A_X(t)-A_R(t)\|
\leq 2|X|\,\|A_X\|e^{-\mu(vt+l)}(e^{2kst}-1).
$$
The factor $\min(|X|,|C|)$ was bounded by $|X|$; there is no sum over sites and hence no volume-dependent prefactor.
Choose
$$
\boxed{v=\frac{4ks}{\mu}.}
$$
For $t,l\geq0$, $e^{2kst}-1\leq2kst\,e^{2kst}$, and therefore
$$
\|A_X(t)-A_R(t)\|
\leq \mu vt\,|X|\,\|A_X\|e^{-2kst-\mu l}
\leq\boxed{\mu vt\,|X|\,\|A_X\|e^{-\mu l/2}.}
$$
Thus $A_{X(l)}(t)=A_R(t)$ is the desired <Lieb-Robinson localization by Haar twirling>. The displayed choice assumes the usual positive constants $\mu,s$; if the interaction bound has $s=0$, the error is identically zero and any positive $v$ works. For negative times the same argument uses $|t|$.
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