Showing these two metrics are equivalent Let $(X,D)$ and $(X,d)$ both be metric spaces on the set $X \subset \mathbb{R}^2$    with,  $$D(\vec{x},\vec{y})=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$
$$d(\vec{x},\vec{y})=|x_2-x_1| + |y_2-y_1|$$
Show that the metrics $D$ and $d$ are equivalent to each other.  
The definition of equivalence being used the following, 
Two metrics $d_1$ and $d_2$ of a set $X$ are equivalent if and only if for any $p \in X$ the following holds for some $\delta_1,\delta_2,\epsilon >0$
$$ N_{d_1}(p, \delta_1) \subset N_{d_2}(p, \epsilon)$$
$$N_{d_2}(p,\delta_2) \subset N_{d_1}(p,\epsilon)$$
My attempt:  Looking at $N_D(0,1)$ and $N_d(0,1)$ helped me gain some intuition as to what I need to do as  $N_D(0,1)$ represents an open unit circle, and $N_d(0,1)$ represents a diamond completely enclosed within it.  Thus showing that $N_d(x,\delta) \subset N_D(x,\epsilon)$ isn't to bad.
What I really need some help on is showing the converse is true.  Through some geometry I think that setting $\delta=\sqrt{2}\,\epsilon$ will work, i'm just having a tough time actually showing this to be the case.  Any help or suggestions are appreciated, Thanks!!
 A: The exact condition for equivalence you want is:
$$\forall p \in X: \forall \epsilon>0: \left(\exists \delta_1>0: N_{d_1}(p, \delta_1) \subseteq N_{d_2}(p,\varepsilon)\right) \land \left(\exists \delta_2 > 0 : N_{d_2}(p, \delta_2) \subseteq N_{d_1}(p,\varepsilon)\right)$$ 
In this case we can prove a global inequality that implies it.
For all $x,y \in \mathbb{R}^2$:
$D(x,y)^2 = (x_1 - y_1)^2 + (x_2 - y_2)^2 \le 2\max(|x_1- y_1| ,|x_2- y_2|)^2$,
so that $D(x,y) \le \sqrt{2}\max(|x_1 - y_1|,|x_2 -y_2|)\le \sqrt{2}(|x_1 - y_1| + |x_2 - y_2|)  =\sqrt{2}d(x,y)$. So
$$D(x,y) \le \sqrt{2}d(x,y)$$
On the other hand $d(x,y) = |x_1 - y_1| + |x_2 - y_2| \le 2\max(|x_1 - y_1|,|x_2 - y_2|)$ and as clearly $\max(|x_1 - y_1|,|x_2 - y_2|)^2 \le D(x,y)^2$, taking roots and combining we get  
$$d(x,y) \le 2D(x,y)$$
So for $d_1 = D, d_2 = d$ in the condition:
for $\varepsilon > 0$ we can take $\delta_1 = \frac{\varepsilon}{\sqrt{2}}$ and $\delta_2 = \frac{\varepsilon}{2}$ for all $p$.
(check the two ball inclusions)
