= Solution
For $0<r<1$, $r\log r<0$. The printed upper bound is negative whenever $\Omega\ne0$, so it cannot bound a nonnegative difference. \b[The correct small-distance modulus is $r\log(1/r)$], and the radius in the hint must also be positive.
Write $h=|x_1-x_2|\leq e^{-1}$, $x_3=(x_1+x_2)/2$, and $M=h\log(1/h)$. Then $h\leq M\leq e^{-1}$. Denote the <planar vorticity velocity kernel> by $K$. It satisfies $|K(z)|\leq C/|z|$ and $|DK(z)|\leq C/|z|^2$. Split the <integral> at radii $M$ and $1$ about $x_3$.
On $|y-x_3|\geq1$, every point of the segment joining $x_1$ and $x_2$ remains at least $|y-x_3|-h/2\geq(1-h/2)|y-x_3|$ from $y$. The <mean value theorem> bounds the kernel difference by $Ch/|y-x_3|^2$, giving a contribution at most $Ch\|\Omega\|_1$.
On $M\leq|y-x_3|\leq1$, the same estimate holds because $h/2\leq M/2$. <Polar integration> gives
$$
\int_{M\leq|y-x_3|\leq1}
|K(x_1-y)-K(x_2-y)|\,|\Omega(y)|\,dy
\leq Ch\|\Omega\|_\infty\int_M^1\frac{dr}{r}
\leq Ch\log(1/h)\|\Omega\|_\infty.
$$
Here $\log(1/M)\leq\log(1/h)$ because $M\geq h$.
On $|y-x_3|<M$, use the sum of the two kernel magnitudes instead of their derivatives. This disk lies inside a disk of radius $M+h/2\leq3M/2$ about either $x_i$. Hence its contribution is bounded by $CM\|\Omega\|_\infty$. Combining the three regions yields
$$
\boxed{|U[\Omega](x_1)-U[\Omega](x_2)|
\leq C(\|\Omega\|_1+\|\Omega\|_\infty)
h\log(1/h),\quad 0<h\leq e^{-1}.}
$$
At coincident points the difference is zero. At larger distances use the boundedness from part (d). A convenient global <log-Lipschitz modulus> is
$$
\mu(r)=
\begin{cases}r(1-\log r),&0<r\leq1,\\1,&r\geq1,\end{cases}
\qquad \mu(0)=0,
$$
giving $|U[\Omega](x)-U[\Omega](y)|\leq C(\|\Omega\|_1+\|\Omega\|_\infty)\mu(|x-y|)$. This is positive, continuous and nondecreasing. It also proves <continuity> of $U[\Omega]$.
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