This is more of a personal question about neighborhoods in $\mathbb R^n$.
Consider some neighborhood $N_{1 \over 12}({\bf 0})$ in $\mathbb R^3$. For any $x\in N_{1 \over 12}({\bf 0})$, are $|x_1|,|x_2|,|x_3| < {1\over 12}$? (where $x = (x_1, x_2, x_3)^T$)
Unfortunately my experience of real analysis in multiple dimensions is limited, but I say that yes, it is true that $|x_1|, |x_2|, |x_3| < 1/12$.
From regular analysis, I can recall that $N_\epsilon(a) = \{x \in \mathbb R : |x - a| < \epsilon\}$, so I assume that in $\mathbb R^n$, the extension is that $N_\delta({\bf x}_0) = \{{\bf x}\in\mathbb R ^n : \|{\bf x - x_0}\| < \delta\}$, is this correct? I am having some difficulty locating information about this in $\mathbb R^n$.
So then it should follow that the norm of any $x\in\mathbb R^3$ will be less than ${1 \over 12}$, and the only way this is possible is if $|x_1|, |x_2|, |x_3| < {1\over 12}$. For example, consider the vector $${\bf u} = \pmatrix{{1\over12}\\ 0 \\ 0} \implies \|{\bf u}\| = \sqrt{1 \over 144} = {1 \over 12}.$$ Hence ${\bf u} \not \in N_{1\over 12}({\bf 0})$. Were the zero values suddenly greater than $0$, then the magnitude would be greater than $1/12$ and also not in $N_{1 \over 12}({\bf 0})$.
Recall that $||\bf{x -y} || = \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2 + (x_3-y_3)^2}$.
So what you want to prove is that for all real numbers $x_1,x_2,x_3$ \begin{align} \sqrt{x_1^2 + x_2^2 + x_3^2} < \frac{1}{12} \implies |x_i| < \frac{1}{12} &\text{ for } i = 1, 2,3 \end{align}
Now $\sqrt{\cdot}$ is an increasing function so $|x_1| =\sqrt{ x_1^2 } \leq \sqrt{x_1^2 + x_2^2 + x_3^2} < \frac{1}{12}$ as desired.