We're talking about the $\Bbb{R^n}$ case here. This is the part of the proof, were we are choosing an appropriate partition so that we can arrive at the Riemann-condition for integrability.
Let $I=[a_1,b_1]\times \dots [a_n,b_n]$. Choose an even partition for $I$, such that $P_\epsilon=\{a_1,a_1+\frac{b_1-a_1}{N},\cdots,b_1\}\times \cdots\{a_n,a_n+\frac{b_n-a_n}{N},\cdots ,b_n\}$, where $N$ is chosen such that $\max_{1\leq j \leq n}\frac{b_j - a_j}{N}<\frac{\delta}{n}$ $\forall x,y\in I_\hat{j}$, and $\hat{j} = (j_1,...,j_n)$. The part I don't understand is this $$\vert\vert \hat{x} -\hat{y} \vert\vert \leq\sum_{j=1}^n\vert x_j-y_j\vert\leq\sum_{j=1}^n\frac{b_j-a_j}{N}<n\frac{\delta}{n}=\delta$$
How do we arrive at the second inequality?
This is only true if the points $\hat{x} = (x_1,\ldots,x_n)$ and $\hat{y} = (y_1,\ldots,y_n)$ are in the same subrectangle, that is there exist $k_1, k_2, \ldots, k_n\in \{1,2,\ldots,n\}$ such that
$$\hat{x}, \hat{y} \in \left[a_1+ \frac{(b_1-a_1)(k_1-1)}{N}, a_1+ \frac{(b_1-a_1)k_1}{N}\right] \times \ldots \times \left[a_n+ \frac{(b_n-a_n)(k_n-1)}{N}, a_n+ \frac{(b_n-a_n)k_n}{N}\right] $$
In that case, both of the components $x_j,y_j$ lie in the same interval of length $\frac{b_j-a_j}{N}$, and we have
$$|x_j - y_j| \leqslant \frac{b_j-a_j}{N}$$