What makes the internal resistance of a battery at microscopic level? What makes the internal resistance of a battery at the microscopic level? 
And why does voltage drop when in a circuit compared to the open circuit voltage? 
I think that the field between the plates should change, lower voltage means weaker electric field.
And the electric field is generated by excess charges in the two plates. But I don't think that there is less excess charges, because even if one more charge leaves the electrode then it get's remplaced by an other, it will make ,after a long, time the electrodes neutral.
My suggestion is that there is some sort of other field inside the cell that counter acts the electrodes plates when there is a current flowing this field must get stronger.
 A: 
What makes the internal resistance of a battery at the microscopic
  level?

There is no simple answer due to the many different types of batteries and technologies. So the following is only an overview.
There are generally two categories of internal resistance of a battery. 
One has to do with the resistivity of the materials of the internal components (e.g electrodes) that conduct charge from and to the terminals. 
The other has to do with the mobility of the ions moving internally between the electrodes through the electrolyte and the medium separating the electrodes.
Other factors include, but are not limited to, battery size, the particular battery chemistry, temperature and age.

And why does voltage drop when in a circuit compared to the open
  circuit voltage? 
  My suggestion is that there is some sort of other field inside the
  cell that counter acts the electrodes plates when there is a current
  flowing this field must get stronger.

I think you are getting close, but I think that what you are referring to as "some sort of field inside the cell" is in reality what we call the emf (electromagnetic force) or voltage that the battery generates inside the battery.
The internal battery voltage is in series with its internal resistance. When current is delivered to the circuit there is a voltage drop across the internal resistance. The terminal voltage is then the emf minus the internal voltage drop. When no current is delivered (open circuit), the terminal voltage is the same as the emf.
The figure below illustrates the point. If $R_L$ is infinite (meaning we have an open circuit) the voltage across the battery terminals equals it internal voltage, or emf. For any finite value of $R_L$ current will occur. That means a  voltage drop across its internal resistance $R_b$. The voltage available at the terminals will equal the emf minus the voltage drop across $R_b$.
Hope this helps 

