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.



Infimum of a finite number of distances in the plane
It can be zero.


Take the standard metric on $\mathbb R^3$ and the one given in [this example](https://mathoverflow.net/questions/125283/on-lipschitz-embeddability-of-certain-compact-metric-spaces-into-mathbbrn/125295#125295) 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 $d(x,y)=0$. 


