Are d1 and lift metric equivalent distances? I need help proving that $d1\not\equiv d$, where d and d1 are defined as follows:
\begin{equation}
     \label{eq:aqui-le-mostramos-como-hacerle-la-llave-grande}
     d(x,y) = \left\{
        \begin{array}{ll}
   |x_{2}-y_{2}|      & \mathrm{if \ }x_{1}=y_{1} \\
  |x_{2}|+|x_{1}-y_{1}|+|y_{2}|    &  \mathrm{otherwise\ }
        \end{array}
      \right.
   \end{equation}
and $d_1=|x_1-y_1|+|x_2-y_2|$
$x,y\in \mathbb{R}^2$ where $x=(x_1,x_2)$ and $y=(y_1,y_2)$,
I have that
$d_1(x,y)\leq d(x,y) $, $B_d(x;\epsilon)\subseteq B_{d_1}(x;\epsilon)$ and $\tau_u\subseteq \tau_d$ where $\tau_u$ is the usual topology induced by $d_1,d_2,d_\infty$ on $\mathbb{R}^2$and $\tau_d$ the topology induced by d on $\mathbb{R}^2$.
I need to prove that HOWEVER $d_1$ and $d$ are not equivalent.
Thanks.

$D=\{0\}\times \{y:1<y<3\}$ is the open $d$-ball of radius $1$ centered at the point $(0,2).$ It is not open in the $d_1$-metric. 
For any $p=(x,y)\in \Bbb R^2$ and any $r>0$ the open $d_1$-ball $B_{d_1}(p,r)$ contains points $(x',y')$ with $x'\ne x. \;$ E.g. $(x+r/2,y)\in B_{d_1}(p,r). $   So no non-empty open $d_1$-ball is a subset of $D.$ And $D$ is not empty. So $D$ cannot be a union of open $d_1$-balls.
The metric $d$ has been called the River Metric (e.g. in General Topology by R. Engelking): For real $x,x'$ with $x\ne x',$ the sets $\{x\}\times \Bbb R$ and $\{x'\}\times \Bbb R$ are separated by mountains, except for the river, which is  $\Bbb R\times \{0\}.$ To get from $(x,y)$ to $(x',y')$ you must travel to $(x,0)$ and along the river to $(x',0)$ and then to $(x',y').$
