Resistance between two points on a conducting surface Suppose we have a cylindrical resistor, with resistance given by $R=\rho\cdot l/(\pi r^2)$
Let $d$ be the distance between two points in the interior of the resistor and let $r\gg d\gg l$. 
Ie. it is approximately a 2D-surface (a rather thin disk).
What is the resistance between these two points?
Let $r,l\gg d$, (ie a 3D volume), is the resistance $0$ ?

Clarification:
A voltage difference is applied between two points a distance $d$ apart, inside a material with resistivity $\rho$, and the current is measured, the proportionality constant $V/I$ is called $R$. The material is a cylinder of height $l$ and radius $r$, and the two points are situated close to the center, we can write $R$ as a function of $l$, $r$ and $d$, $R(l,r,d)$, for small $d$.
The questions are then:
What is $$ \lim_{r \rightarrow \infty} \lim_{l \rightarrow \infty}  R(l,r,d) $$
What is $$ \lim_{r \rightarrow \infty} \lim_{l \rightarrow 0} R(l,r,d) $$
 A: Potential for 2D problem
Let's start with a 2D disk and try to solve the general problem for infinitesimally flat disk.  I will change notations a bit -- the surface resistance will be $\sigma$ and the radius of the disk will be $a$.
Starting with basic electrodynamics:  
$\vec{j} = -\sigma\frac{\partial u}{\partial \vec{r}},\, div\vec{j}=0\,\Rightarrow\,\Delta u = 0$ with the boundary condition: $\vec{n}\cdot\vec{j} = 0 \Rightarrow \vec{n}\frac{\partial u}{\partial \vec{r}} = 0$
Let's first consider the current $I$ flowing into the surface in the centre and uniformly flowing away from the edges. solution for potential is well known:
$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln r$ 
I use the conformal map $z\to a\frac{z-s}{a^2-s*z}$ to "shift the centre" into the point $s=x_{source}+iy_{source}$. The potential is then:
$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln\left|a\frac{re^{i\phi}-s}{a^2-s^*re^{i\phi}}\right|$
Now I substract the similar potential, with different parameter $d=x_{drain}+iy_{drain}$ to compensate the outgoing flow. Obtaining:
$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln\left|\frac{re^{i\phi}-s}{re^{i\phi}-d}
\cdot\frac{a^2-d^*re^{i\phi}}{a^2-s^*re^{i\phi}}\right|$ or $U(z) = -\frac{I}{2\pi\sigma}\, \ln\left|\frac{z-s}{z-d}
\cdot\frac{a^2-d^*z}{a^2-s^*z}\right|$
This is the harmonic function, satisfying the boundary conditions. You can play here with it.
Interpretation of the solution 
The potential is divergent in points $s$ and $d$. This happens because the resistance is strongly dependend on the microscopic details of the problem. Indeed -- as you get closer to the source -- all your current have to pass through smaller and smaller amount of conductor. And in the limit of infinitely small source you get infinite resistance.
Formulation issue
I admit that while solving I first fixed the current and then found the potential, while you formulated the problem differently -- "set the potential here and there and find the current". But let us use logic:  


*

*Nonzero current leads to infinite voltage: $I\neq0\,\Rightarrow\,\Delta U \to \infty$.

*If $A\Rightarrow B$, then $!B\Rightarrow !A$.

*$\Delta U\mbox{-finite}\,\Rightarrow\,I=0$ 


At finite voltage you'll get zero current or, equivalently, infinite resistance.
What happens in 3D case?
Same thing. Just consider single pointlike source -- and the potential $U\sim\frac{1}{r}$ is divergent. Don't need to go into further details.
"Cutoffs"
In order to move on I introduce the "cutoffs" -- new small (real) quantities $\epsilon_{s,d}$ which denoting "sizes" of the source and the drain. Using them I obtain the voltage:  
$U(d+\epsilon_d)-U(s+\epsilon_s)=\frac{I}{2\pi\sigma}\left[\ln\frac{\epsilon_s}{|s-d|}+\ln\frac{\epsilon_d}{|s-d|} +\ln\left|\frac{a^2-s^*d}{a^2-|s|^2}\cdot\frac{a^2-sd^*}{a^2-|d|^2}\right|\right]$
Scales
Putting together everything above. One can say that in the problem there are  four (or, even five) scales:


*

*Radius of the disk.

*Thickness of the disk. 

*Distance between contacts $|s-d|$

*Sizes of those contacts $\epsilon_{s,d}$


Since you are talking about "points" -- then first we have to take $\epsilon_{s,d}\to0$, right? But if $\epsilon_{s,d}$ is much smaller that any other scale then they introduce divergent contribution into the resistance. And any other detail of the problem becomes irrelevant.  
Therefore, the answer to your question is: The resistance between two points is infinite, whatever the geometry of the problem is. 
A: I think the answer is infinite because of the singular nature of a point. If we assume a steady state situation then the divergence of the potential is zero, by symmetry in a full 2D model the current density J will scale like 1/r. This implies the voltage scales like log(r), which diverges as r goes to zero. In the 3D case its even worse, as J scales as 1/r**2, and thus voltage scales as 1/r, which diverges even faster as r goes to zero. Note that we can compute restance, by fixing the current, and computing the voltage difference. The problem is that the voltage difference doesn't converge in the vicinity of a point current source/sink.
A: The resistance between two electrodes each of radius r0 (not points) separated by a distance of 2S on an infinite plane is given by 
arcosh(s/r0)/pi/resistivity
https://www.physicsforums.com/proxy.php?image=http%3A%2F%2Fimg11.hostingpics.net%2Fpics%2F843563pourforum11.jpg&hash=7b69021109a0ada9b50b7ba16cfd1414
See also thermoconductivity Shape Factors
A: "Resistance between two points" is not a well-defined concept.  If  c  is a curve connnecting  point  A  to point  B, the electromotive force  V  along  c  is a well-defined thing (it's the circulation of the electric vector field along  c), and resistance would be the factor  R  such that  V = RI  when  I  is the intensity flowing through a thin tube borne by  c.  But it's clear that intensity  depends on the cross section of this virtual tube, a non-defined entity here.   So we have no proper definition of "resistance  R  between  A  and  B" that would allow us to compute it.
A: I'll tackle the 3D case. I am using the SI system. It should be noted that the electrical resistance of an electrical element (in given case a homogeneous medium) measures its opposition to the passage of an electric current (in given case direct current). The resistance of a homogeneous medium between two electrodes is defined as $$R=\frac{U}{I}=\frac{\rho\epsilon_0}{C}$$ where C is a capacitance between two electrodes.  
Let's start with the assumption that instead of two points we have two conducting tiny spheres with a radius $r_0$. The distance between the centers of the spheres is $d$ and $r_0<<d$. To simplify the calculation let us assume that the charge on the spheres is distributed spherically symmetric. Then $$C=2\pi\epsilon_0r_0$$ Finally, required resistance:  
$$R=\frac{\rho}{2\pi r_0}$$
