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Found the proof! It's done using the integral definition of $T$: $$ T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=0}^n R\left(\tfrac{i}{n}v\right) $$ So for any vectors $X$ and $Y$: \begin{align*} &\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\ &\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=0}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \end{align*} and we only need to prove that each summand is less than $|u-v| |X| |Y|$: \begin{align*} & \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\ &= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\ & \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\ &+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\ &\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\ &= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr| \end{align*} where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this question). Since $X$ and $Y$ are arbitrary, $$ \left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v| $$ in the subordinate norm.

Found the proof! It's done using the integral definition of $T$: $$ T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=1}^n R\left(\tfrac{i}{n}v\right) $$ So for any vectors $X$ and $Y$: \begin{align*} &\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\ &\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \end{align*} and we only need to prove that each summand is less than $|u-v| |X| |Y|$: \begin{align*} & \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\ &= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\ & \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\ &+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\ &\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\ &= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr| \end{align*} where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this question). Since $X$ and $Y$ are arbitrary, $$ \left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v| $$ in the subordinate norm.

Found the proof! It's done using the integral definition of $T$: $$ T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=0}^n R\left(\tfrac{i}{n}v\right) $$ So for any vectors $X$ and $Y$: \begin{align*} &\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\ &\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=0}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \end{align*} and we only need to prove that each summand is less than $|u-v|$: \begin{align*} & \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\ &= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\ & \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\ &+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\ &\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\ &= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr| \end{align*} where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this question). Since $X$ and $Y$ are arbitrary, $$ \left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v| $$ in the subordinate norm.

Found the proof! It's done using the integral definition of $T$: $$ T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=0}^n R\left(\tfrac{i}{n}v\right) $$ So for any vectors $X$ and $Y$: \begin{align*} &\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\ &\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=0}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \end{align*} and we only need to prove that each summand is less than $|u-v| |X| |Y|$: \begin{align*} & \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\ &= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\ & \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\ &+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\ &\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\ &= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr| \end{align*} where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this question). Since $X$ and $Y$ are arbitrary, $$ \left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v| $$ in the subordinate norm.

Found the proof! It's done using the integral definition of $T$: $$ T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=0}^n R\left(\tfrac{i}{n}v\right) $$ So for any vectors $X$ and $Y$: \begin{align*} &\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\ &\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=0}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \end{align*} and we only need to prove that each summand is less than $|u-v|$: \begin{align*} & \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\ &= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\ & \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\ &+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\ &\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\ &= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr| \end{align*} where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this answer). Since $X$ and $Y$ are arbitrary, $$ \left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v| $$ in the subordinate norm.

Found the proof! It's done using the integral definition of $T$: $$ T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=0}^n R\left(\tfrac{i}{n}v\right) $$ So for any vectors $X$ and $Y$: \begin{align*} &\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\ &\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=0}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \end{align*} and we only need to prove that each summand is less than $|u-v|$: \begin{align*} & \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\ &= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\ & \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\ &+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\ &\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\ &= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr| \end{align*} where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this question). Since $X$ and $Y$ are arbitrary, $$ \left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v| $$ in the subordinate norm.