Infimum of a finite number of distances in the plane Suppose one has a finite number of distances $d_1,\ldots,d_k$ on the Euclidean plane all of which metricize the usual Euclidean topology.
Define for each pair of points $x$ and $y$ in the plane
$$d(x,y) = \inf\left\lbrace d_{i_1}(x_0,x_1) + \cdots d_{i_l}(x_{l-1},x_l) \right\rbrace$$
where the infimum is taken over all finite sequences with $x_0 = x,x_1,\ldots,x_l = y$ and all possible choices of $i_1,\ldots, i_l$.
Question: Is it possible for $d(x,y)$ to be $0$ for a pair of distinct points $x \neq y$?
Also, I'm interested in the same question replacing $\mathbb{R}^2$ by an arbitrary Polish space.
 A: The infimum can be zero as pointed out by Anton Petrunin.  Here's a construction on the interval $[0,1]$.
Consider a sequence of piecewise linear functions $f_n:[0,1] \to [0,1]$ each of which is strictly increasing defined by (see figure):  


*

*$f_0(x) = x$ for all $x$.

*$f_{n+1}$ coincides with $f_n$ on diadic numbers of denominator $2^{-n}$.

*$f_{n+1}\left(\frac{2k+1}{2^{n+1}}\right) = f_n\left(\frac{k}{2^n}\right) + \frac{1}{3^{n+1}}$ for integer $k$.


Let $f = \lim\limits_{n \to +\infty}f_n$ and $d_1$ be the pullback of the standard Euclidean distance on $[0,1]$ under $f$.   One has that
$$d_1(0,1/2) = \frac{1}{3}$$
$$d_1(0,1/4) = d_1(1/2,3/4) = 1/9$$
$$d_1(0,1/8) = d_1(1/4,3/8) = d_1(1/2,5/8) = d_1(3/4,7/8) = 1/27$$
etc...
One can construct in similar fasion a distance $d_2$ satisfying 
$$d_2(1/2,1) = \frac{1}{3}$$
$$d_2(1/4,1/2) = d_2(3/4,1) = 1/9$$
$$d_2(1/8,1/4) = d_2(3/8,1/2) = d_2(5/8,3/4) = d_2(7/8,1) = 1/27$$
etc...
By considering diadic partitions one sees that the infimum is $0$ when one can use the distances $d_1$ and $d_2$.


A: It can be zero.
Take the standard metric on $\mathbb R^3$ and the one given in this example by S. Ivanov.
Below I give simplification of his example which works in your case.
Simplified example.
Choose two points $x$ and $y$ 
in $\mathbb R^2$ and construct a sequence of arcs $\gamma_n$ between them.
For each $n$ choose small $\varepsilon_n>0$, so that $\varepsilon_n\to 0$ very fast.
Choose disjoint $\varepsilon_n$-intervals on $\gamma_n$, so that it cover all $\gamma_n$ except set of lenght $\varepsilon_n$.
Construct metric $d_1$ so that it is very cheap to go along each such interval, but compensate it by making very expancive to get to such interval, so you can not use it as a shortcut.
Let $d_2$ be the Euclidean metric. 
Then the $d$-length of $\gamma_n$ has order of $\varepsilon_n$. In particular $d(x,y)=0$. 
A: This is possible even on the real line.
There is a strictly increasing continuous function $f:[0,1]\to\mathbb R$ whose derivative is zero almost everywhere. It is a suitable sum of a series of Cantor functions. See, for example, Gelbaum and Omsted, "Counterexamples in analysis" (1964), Chapter 8, Example 30.
This function has the following property: for every $\varepsilon>0$ there is a collection of disjoint intervals $[a_i,b_i]\subset[0,1]$ with total length greater than $1-\varepsilon$ and total variation of $f$ less than $\varepsilon$:
$$
\sum (b_i-a_i)>1-\varepsilon, \qquad \sum (f(b_i)-f(a_i)) < \varepsilon .
$$
Now define metrics $d_1$ and $d_2$ on $[0,1]$ as follows: $d_1$ is the standard metric and $d_2$ is the pull-back of the standard metric by $f$, i.e. $d_2(x,y)=|f(x)-f(y)|$. You can go from 0 do 1 through points $a_1,b_2,a_2,b_2,\dots$ using the distance $d_2$ between $a_i$ and $b_i$ and $d_1$ between $b_i$ and $a_{i+1}$. Thus $d(0,1)<\varepsilon$ for every $\varepsilon>0$.
A: Unless I'm missing something, the answer is almost trivially no:
For all i, $d_i$ takes values in $\mathbb{R}^+$, so the $d(x,y)=0$ if and only if there is a decomposition such that for all $j$, $d_{i_j}(x_{j-1},x_j)=0$, which happens if and only if $x_{j-1}=x_j$. Therefore, if $d(x,y)=0$, then $x=x_0=x_1=\dots=x_l=y$ (and vice-versa).
