Infimum of a finite number of distances in the plane

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 $\varepsilon>0$.

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):

1. $f_0(x) = x$ for all $x$.
2. $f_{n+1}$ coincides with $f_n$ on diadic numbers of denominator $2^{-n}$.
3. $f_{n+1}\left(\frac{2k+1}{2^{n+1}}\right) = f_n\left(\frac{k}{2^n}\right) + \frac{1}{3^{n+1}}$ 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 $d_2$.

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 $d(x,y)=0$.