Let $f:\mathbb R^d \to \mathbb R, (t, x) \mapsto \|x\|^2$. Then $f$ is smooth and time-independent. The derivatives of $f$ are as follows:
$$\frac{\partial}{\partial x_i} f(x) = 2x_i, \quad \frac{\partial^2}{\partial x_i\partial x_j} f(x) = 2\delta_{ij}$$
By Itô's formula:
\begin{align}
\mathrm d f(X_t-Y_t) &= \sum_{i=1}^d \frac{\partial f}{\partial x_i} (X_t-Y_t) \mathrm d(X^i_t-Y^i_t)\\
&+ \frac 12 \sum_{i,k=1}^d \frac{\partial^2 f}{\partial x_i\partial x_k} (X_t-Y_t) \sum_{j=1}^r (\sigma_{ij}(t,X_t)-\sigma_{ij}(t,Y_t))(\sigma_{kj}(t,X_t)-\sigma_{kj}(t,Y_t)) \mathrm dt\\
&= \sum_{i=1}^d 2(X^i_t-Y^i_t) (b^i(t,X_t)-b^i(t,Y_t)) \mathrm dt\\
&+ \sum_{i=1}^d 2(X^i_t-Y^i_t) \sum_{j=1}^r (\sigma_{ij}(t,X_t)-\sigma_{ij}(t,Y_t)) \mathrm dW^j_t\\
&+ \frac 12 \sum_{i=1}^d 2\sum_{j=1}^r (\sigma_{ij}(t,X_t)-\sigma_{ij}(t,Y_t))^2 \mathrm dt\\
&= 2\langle X_t-Y_t, b(t,X_t)-b(t,Y_t)\rangle \mathrm dt\\
&+ 2\sum_{i=1}^d \sum_{j=1}^r (X^i_t-Y^i_t)(\sigma_{ij}(t,X_t)-\sigma_{ij}(t,Y_t)) \mathrm dW^j_t\\
&+ \|\sigma(t,X_t)-\sigma(t,Y_t))\|^2 \mathrm dt
\end{align}
Here, $\|\cdot\|$ denotes the Frobenius norm. Furthermore:
$$f(X_0-Y_0) = f(\Xi-\Xi) = f(0) = 0$$
Integrating $\mathrm df(X_t-Y_t)$ from $0$ to $t$ now yields the following identity:
\begin{align}
\|X_t-Y_t\|^2 &= 2\int_0^t \langle X_t-Y_s, b(s,X_s)-b(s,Y_s)\rangle \mathrm ds\\
&+ 2\sum_{i=1}^d \sum_{j=1}^r \int_0^t (X^i_s-Y^i_s)(\sigma_{ij}(s,X_s)-\sigma_{ij}(s,Y_s)) \mathrm dW^j_s\\
&+ \int_0^t \|\sigma(s,X_s)-\sigma(s,Y_s))\|^2 \mathrm ds
\end{align}