How to estimate the difference between two Ito diffusions?

Question

Suppose $𝑏:\mathbb R^d \to \mathbb R^d, \sigma:\mathbb R^d \to \mathbb R^{d\times d}$ are measurable functions and satisfy \begin{equation*} 2\langle 𝑥−𝑦,𝑏(𝑥)−𝑏(𝑦)\rangle +\|\sigma(𝑥)−\sigma(𝑦)\|^2 \le 𝐾|𝑥−𝑦| \rho(|𝑥−𝑦|),\ \ \ \ {(1)}\end{equation*} for all $𝑥,𝑦 \in \mathbb R^d$, where $\rho:[0, \infty)→[0,\infty)$ is some appropriate function. In addition, assume that the SDE $$𝑑𝑋(𝑡)=𝑏(𝑋(𝑡))𝑑𝑡+ \sigma(𝑋(𝑡))𝑑𝑊(𝑡)$$ is well-posed in the strong sense.

We now consider the coupled process $$\begin{aligned} & 𝑑𝑋(𝑡)=𝑏(𝑋(𝑡))𝑑𝑡+\sigma(𝑋(𝑡))𝑑𝑊(𝑡), \\ & 𝑑𝑌(𝑡)=𝑏(𝑌(𝑡))𝑑𝑡+\sigma(𝑌(𝑡))𝑑𝑊(𝑡),\end{aligned}$$ where $𝑊$ is a $𝑑$-dimensional standard Brownian motion. Is it true that for all $t\geq 0$ the following estimate is true? \begin{equation*} 𝑑|𝑋(𝑡)−𝑌(𝑡)|≤\rho(|𝑋(𝑡)−𝑌(𝑡)|)𝑑𝑡+\left\langle \frac{𝑋(𝑡)−𝑌(𝑡)}{|𝑋(𝑡)−𝑌(𝑡)|},(\sigma(𝑋(𝑡))−\sigma(𝑌(𝑡)))𝑑𝑊(𝑡)\right\rangle.\ {(2)}\end{equation*}

Using Ito's formula, we have $$ 𝑑|𝑋(𝑡)−𝑌(𝑡)|^2=(2\langle 𝑋(𝑡)−𝑌(𝑡),𝑏(𝑋(𝑡))−𝑏(𝑌(𝑡))\rangle + \|\sigma(𝑋(𝑡))−\sigma(𝑌(𝑡))\|^2)𝑑𝑡+2\langle 𝑋(𝑡)−𝑌(𝑡),(\sigma(𝑋(𝑡))−\sigma(𝑌(𝑡)))𝑑𝑊(𝑡)\rangle.$$ Using (1), and applying Ito's formula to the process $|𝑋(𝑡)−𝑌(𝑡)|^2$ using the function $𝑓(𝑟)=𝑟^{1/2}, 𝑟\ge 0,$ we can formally derive (2). But there is a problem, as $𝑓$ is not differentiable at 0. But I don't know how to rigorously prove or disprove (2).

Thanks a lot for your help. (I posted this question in Stack Exchange, but no answer yet).