= Solution
Let $g_t=P_{1-t}(\nabla f)(X_t)$. The <triangle inequality> for the conditional average gives
$$
\lVert g_t\rVert\le P_{1-t}(\lVert\nabla f\rVert)(X_t)\le K.
$$
The independent coordinate <Brownian motions> have zero cross-variation, so the <quadratic variation of a stochastic integral> gives
$$
\boxed{[M]_1=\int_0^1\lVert g_t\rVert^2\,dt\le K^2.}
$$
The <Dambis-Dubins-Schwarz theorem> says that a <continuous local martingale> $N$ starting at zero whose <quadratic variation> diverges admits a <Brownian motion> $B$, with the inverse-clock filtration, such that $N_t=B_{[N]_t}$. If the total clock is finite, one may enlarge the probability space and append independent <Brownian motion> after the clock terminates. This gives the same identity up to the original lifetime.
Apply this theorem to $N=M-M_0$. The clock bound implies the event inclusion
$$
\{|M_1-M_0|>r\}\subseteq\left\{\sup_{0\le s\le K^2}|B_s|>r\right\}.
$$
The latter event is contained in the union of the upper and lower crossing events. Symmetry of <Brownian motion> gives
$$
\boxed{\mathbb P(|M_1-M_0|>r)\le2\mathbb P\left(\sup_{0\le s\le K^2}B_s>r\right).}
$$
This uses only an event inclusion; independence between the <Brownian motion> and its random clock is neither assumed nor needed.
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