= Solution
Literally, the printed hypothesis with positive $C$ is impossible for pairs at distance less than one, even when $U$ is constant. Use the corrected <log-Lipschitz modulus> $\mu$ from part (e). Local existence for the continuous finite-dimensional <vector field> follows from the <Peano existence theorem>. Boundedness gives $|X(t)-X(s)|\leq\|U\|_\infty|t-s|$. Thus a trajectory cannot escape to infinity in finite time; at a finite endpoint it has a limit, and local existence there extends it. \b[Solutions exist globally.]
For uniqueness, let $X,Y$ have the same initial point and set $\delta(t)=|X(t)-Y(t)|$. This <absolutely continuous function> satisfies, almost everywhere,
$$
\delta'(t)\leq C\mu(\delta(t)),\qquad \delta(0)=0.
$$
This is the <Osgood uniqueness criterion>, rather than the usual linear <Gronwall inequality>, because the velocity need not be <Lipschitz continuous>. To see the argument directly, put $w_\varepsilon=\delta+\varepsilon$. As long as $w_\varepsilon\leq1$, monotonicity of $\mu$ gives
$$
w_\varepsilon'\leq Cw_\varepsilon(1-\log w_\varepsilon).
$$
For $z_\varepsilon=1-\log w_\varepsilon$, this becomes $z_\varepsilon'\geq-Cz_\varepsilon$. Integration gives
$$
w_\varepsilon(t)\leq
\exp\left(1-(1-\log\varepsilon)e^{-Ct}\right).
$$
For any fixed finite interval, this upper bound tends uniformly to zero as $\varepsilon\downarrow0$. Taking $\varepsilon$ small prevents a first exit through $w_\varepsilon=1$, so the bound remains valid throughout that interval. Thus $\delta=0$. Backward time has the same proof. \b[Every initial point has exactly one global trajectory.]
The decisive fact is $\int_0^1dr/\mu(r)=\infty$. For different nearby initial points, the same comparison gives
$$
|X(t)-Y(t)|\leq e^{1-e^{-Ct}}|X(0)-Y(0)|^{e^{-Ct}}
$$
until the separation reaches one. This quantitative continuous dependence makes the flow a homeomorphism, although it need not be a smooth <diffeomorphism>.
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