Show that $d_V$ is a metric Problem: 
For points $p = (p_1, p_2)$ and $q = (q_1, q_2)$ in $\mathbb{R}^2$ define: 
$d_V(p,q) = \begin{cases}1 & p_1\neq q_1 \ or\ |p_2 - q_2|\geq 1  \\ |p_2 - q_2| & p_1= q_1 \ and\ |p_2 - q_2|< 1 \end{cases}$
Show that $d_V$ is a metric. 
My Unfinished Solution: 
So there are three states: 
$1- \ p_1= q_1 \ and\ |p_2 - q_2|\geq 1 \\ 2- \ p_1= q_1 \ and\ |p_2 - q_2|< 1 \\ 3- \ p_1\neq q_1$ 
The first and second criterion of a metric space are easy to prove. But I can't gather all the three above-mentioned states for $(p,q,r)$ to prove metricness of $d_V$  for the third criterion, i.e., $d(p, q) + d(q, r) \geq d(p, r)$ for all $p,q,r \in \mathbb{R}^2$.
Thank you. 
 A: $\forall p,q : \ d_V(p,q) = |p_2 - q_2| \mathrm \ or \  1 \  $ and $\ d_V(p,q) \le 1$ 
Specially,if $ d_V(p,q) =1$ then $d_V(p,q)\le |p_2 - q_2|$
CASE I:  $d_V(p,r)=|p_2 - r_2|\ \le |p_2 - q_2|+|q_2 - r_2|$ 
(Remark: $|p_2 - r_2|\ \le |p_2 - q_2|+|q_2 - r_2|$ is from basic inequation in R: $|a+b|\le |a|+|b|$)
$ \ $
If $|p_2 - q_2|<1$ , $|q_2 - r_2|<1,p_1=q_1,q_1 = r_1$
Then $d_V(p,q)+d_V(q,r)= |p_2 - q_2|+|q_2 - r_2|\ge d_V(p,r)$ 
$ \ $
Other cases, at least one of equations :$\ d_V(p,q) =1 $ , $d_V(q,r)=1$ holds. 
Hence, $d_V(p,q)+d_V(q,r)\ge 1\ge  d_V(p,r)$
$ \ $
(Remark: Follow the above argurment, get that $d_V(p,q)+d_V(q,r) = |p_2 - q_2|+|q_2 - r_2| $ or $ \ge 1$)
CASE II: $d_V(p,r)=1 \le |p_2 - r_2|\le |p_2 - q_2|+|q_2 - r_2|$
By the similiar arguement did in CASE I,find $d_V(p,q)+d_V(q,r)\ge  d_V(p,r)$
A: Suppose we have three distinct points $x_1 = (p_1,q_1), x_2 = (p_2, q_2), x_3 = (p_3, q_3)$ in the plane (in proving the triangle inequality we can always assume that the three points are distinct; when two or more are equal, it already follows from the other axioms)
We want to show $d_V(x_1, x_3) \le d_V(x_1, x_2) + d_V(x_2, x_3)$.
It's clear from the definition that $d_V(x_1, x_3) \le 1$ always (as is any value of $d_V$). So we can assume (or we are done) that none of the two distances on the right hand side is $1$ and their sum is $<1$ as well. This can only happen if $p_1 = p_2$ and $d_V(x_1,x_2) = |q_1 - q_2| < 1$ and $p_2 = p_3$ and $d_V(x_2, x_3) = |q_2 - q_3| < 1$. 
So in particular, $p_1 = p_3$. We know that $|q_1 - q_3| \le |q_1 - q_2| + |q_2 - q_3| = d_V(x_1, x_2) + d_V(x_2, x_3) < 1$ (by assumption), using in the first step the usual inequality for $|\cdot|$, and so $d_V(x_1, x_3) = |q_1 - q_3| \le  d_V(x_1, x_2) + d_V(x_2, x_3)$ and we are done.
