You have $n-x_i \geq z >0$ and $n+x_i \geq z >0$ for all $i$. Furthermore, you have $|y_i-x_i| < z$, which is equivalent to $-z < y_i-x_i < z$.
Then you have $n-y_i = n-x_i +y_i-x_i \geq z + y_i-x_i > z -z= 0$. Similarly, you have $n+y_i = n+x_i - (y_i-x_i) \geq z - (y_i-x_i) > z -z = 0$. Hence $y_i \in (-n,n)$ for all $i$. It follows that $y \in I_n$.
NOTE:
I need to elaborate on a point above. In the statement $y\in N_z(x)$, I was taking the neighborhood to be $\{y \ | \ |y_i-x_i| < z \in \forall i \}$. This is a perfectly valid neighborhood (using the infinity norm $\|\cdot\|_{\infty}$), but may not agree with your norm. A more usual norm would be the 2-norm $\|x\|_2 = \sqrt{\sum_i x_i^2}$, in which case you would need to adjust the neighborhood slightly, since the relevant bound is $\|x\|_2 \leq \sqrt{n} \|x\|_{\infty}$. If you are using the 2-norm, then you should choose $y \in N_{\frac{z}{\sqrt{n}}} (x)$, in which case it will then be true that $-z < y_i-x_i < z$.