$d\left(\left(x_1,x_2\right),\left(y_1,y_2\right)\right)=|x_1-x_2|+|x_1-y_1|+|y_1-y_2|$ : complete? Define the $\Bbb R^2$ metric
$$
d\left( x,y \right) = \begin{cases}
\left|x_2-y_2\right| &, x_1 = y_1 &&\text{(d1)}\\
\left|x_1-x_2\right|+\left|x_1-y_1\right|+\left|y_1-y_2\right| &, x_1 \ne y_1 
&& \text{(d2)}
\end{cases}
$$
where $x=\left(x_1, x_2 \right) \in \Bbb R^2$.
Is $d$ complete? 
I think yes, but I'm stuck in the proof.
Let $\left\{ x_n = \left(x^1_n,x^2_n \right) \right\}$ be $d$-Cauchy.
By the real triangle inequality
$$
\left|x^2_n - x^2_m \right| \le 
\left|x^1_n-x^2_n\right|+\left|x^1_n-x^1_m\right|+\left|x^1_m-x^2_m\right| , \tag{1}
$$
so it's suffice to consider (d2) - Is this correct?
From the middle term in the rhs of (1), $\{x_n^1\}$ is Cauchy in $\Bbb R$, so it converges to a candidate $x^1$.
But then, what is the candidate for $\{x^2_n\}$?
 A: HINT: It will be convenient to let $V_x=\{x\}\times\Bbb R$ for $x\in\Bbb R$ and to let $\Delta=\{\langle x,x\rangle:x\in\Bbb R\}$, the diagonal in $\Bbb R^2$. For any $x=\langle x_1,x_2\rangle\in\Bbb R^2$, $|x_1-x_2|$ is the distance from $x$ to $V_{x_1}\cap\Delta$, the vertical distance from $x$ to the diagonal. Thus, if $x=\langle x_1,x_2\rangle,y=\langle y_1,y_2\rangle\in\Bbb R^2$, either 


*

*$x_1=y_1$, $X,y\in V_{x_1}=V_{y_1}$, and $d(x,y)$ is just the usual Euclidean distance between $x$ and $y$ in $V_{x_1}$;  

*or $x_1\ne y_1$, and $d(x,y)$ is the sum of the vertical distances from $x$ and $y$ to $\Delta$ and the horizontal distance between $V_{x_1}$ and $V_{y_1}$. 


We can make this easier to visualize by looking at an isometric copy of the space. Define
$$\varphi:\Bbb R^2\to\Bbb R^2:\langle x,y\rangle\mapsto\langle x,y-x\rangle\;;$$
$\varphi$ is a vertical shear that holds the origin fixed and pushes $\Delta$ to where the original $x$-axis was. For $x=\langle x_1,x_2\rangle,y=\langle y_1,y_2\rangle\in\Bbb R^2$ let 
$$\rho(x,y)=\begin{cases}
|x_2-y_2|,&\text{if }x_1=y_1\\
|x_2|+|x_1-y_1|+|y_2|,&\text{otherwise}\;.
\end{cases}$$ 
It’s not hard to show that $\varphi$ is a bijection and that for all $x=\langle x_1,x_2\rangle,y=\langle y_1,y_2\rangle\in\Bbb R^2$, $d(x,y)=\rho\big(\varphi(x),\varphi(y)\big)$, so that $\langle\Bbb R^2,d\rangle$ is isometric to $\langle\Bbb R^2,\rho\rangle$, and one is complete iff the other is.
But $\rho$ is easy to visualize: $\rho$ is a variant of the taxicab metric in which the only streets are the verticals $V_x$ and the $x$-axis. The $\rho$-distance from $x=\langle x_1,x_2\rangle$ to $y=\langle y_1,y_2\rangle$ is the ordinary distance if $x$ and $y$ are on the same ‘street’ (i.e., if $x_1=y_1$, and they’re on the same vertical street, or $x_2=y_2=0$, and they’re on the same horizontal street), and in all cases it’s the length of the shortest route from $x$ to $y$ along the streets.
Now show that a $\rho$-Cauchy sequence is either eventually in some $V_x$, in which case convergence is easy, or convergent to the point on the $x$-axis determined by the limit of the $x$-coordinates of the points of the $\rho$-Cauchy sequence. (If you prefer to work directly in $\langle\Bbb R^2,d\rangle$ and give $\rho$ a miss, show that a $d$-Cauchy sequence is either eventually in some $V_x$ or convergent to some point of $\Delta$. You can also, if you like, show that $d$ is a taxicab metric in which the streets are the $V_x$’s and $\Delta$, and distances along $\Delta$ are $\frac1{\sqrt2}$ times the usual Euclidean distances.)
A: The answer is positive. I can give a sketch right now and will expand it with pleasure tomorrow if needed.
First, notice that there is a linear function which is an isometry between $\mathbb R^2$ with your metric and $\mathbb R^2$ with the river metric. The river metric is similar but probably easier to think about:
$$
d\left( x,y \right) = \begin{cases}
|x_2-y_2|, & x_1 = y_1 &&\text{(d1)}\\
|x_2|+|y_2|+|x_1-y_1|, & x_1 \ne y_1 
&& \text{(d2)}
\end{cases}
$$
Then notice that the river metric is complete - for a Couchy sequence coordinates must converge. If the second one converges to a nonzero number, then the first must be constant for $n$ large enough (why?) and we get the convergence. If the second converges to zero, then we also easily prove the existence of the limit.
A: Yes, this is complete. 
But my previous answer was wrong, and it turns out to be easier than I first thought.
First observe that
$$
\max\{|x_1-y_1|,|x_2-y_2|\}\leq d(x,y)
$$
for all $x,y\in\mathbb{R}^2$.
Now let $x^{(n)}=(x_1^{(n)},x_2^{(n)})$ be a Cauchy sequence for $d$.
By the previous observation, it follows that both coordinates are Cauchy in $\mathbb{R}$ so they converge to some 
$$
\lim x_1^{(n)}=x_1\quad\mbox{and}\quad \lim x_2^{(n)}=x_2.
$$
It remains to prove that $x^{(n)}$ converges to $x=(x_1,x_2)$ for $d$.
First case: there exists $N$ such that $x_1^{(n)}=x_1$ for all $n\geq N$. Then
$$
\lim d(x^{(n)},x)=\lim |x_2^{(n)}-x_2|=0.
$$
Second case: there does not exist such an $N$, and so there exists an injective subsequence $(x_1^{(n_k)})$.
Now 
$$
d(x^{(n_k)},x^{(n_l)})=|x_1^{(n_k)}-x_2^{(n_k)}|+|x_1^{(n_k)}-x_1^{(n_l)}|+|x_1^{(n_l)}-x_2^{(n_l)}|.
$$
Letting $k\neq l$ tend to infinity, we get:
$$
0=|x_1-x_2|+0+|x_1-x_2|
$$
so $x_1=x_2$.
Finally, observe that
$$
\lim |x_2^{(n)}-x_2|=0
$$
and
$$
\lim |x_1^{(n)}-x_2^{(n)}|+|x_1^{(n)}-x_1|+|x_1-x_2|=0
$$
so $\lim d(x^{(n)},x)=0$ follows easily.
