Electric potential on different points of a simple circuit
Your schematics have omitted the battery's internal resistance.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. The real battery can be modeled as an ideal voltage source with internal series resistance.


  From what I understand, there exists a potential of sorts between the anode and the cathode terminal of a battery and that I am able to measure this potential with a voltmeter. 
  The potential is real.
  
  I have not been able to test this scenario in real life, since it involves a short circuit, ...


A wise choice.


  ... but from what I understand, when the leads of the voltmeter are placed close to the battery terminals, there should be a reading indicating a presence of voltage. This example is shown in the first picture, and I am assuming that there would be a potential of 12 V present.


Due to the internal resistance of the battery the terminal voltage will collapse to almost zero when a short circuit is applied. I say 'almost' because your best short circuit will have some resistance.


  My question is, as we move the voltmeter test leads closer to one another, would there be an observable change in the potential?


It's simply a matter of the ratio of the internal resistance to that of your short circuit.


  I am making the assumption that as the test leads get closer together, the potential between these two points in space is getting lower and I should be able to measure this change in my voltmeter. 


You are correct. Let's do a quick example.


12 V battery.
Battery short-circuit current is 20 A.
\$ R_{internal} = \frac {V_{OC}}{I_{SC}} = \frac {12}{20} = 0.6 \ \Omega \$ where OC is open-circuit and SC is short-circuit.
Let's put a 0.001 Ω "short" across the terminals.
Terminal voltage will drop to \$ V_T = V_{OC} \frac { R_{SC} }{R_{INT}+R{SC}} = 12 \frac { 0.001} {0.001+0.6} = 0.0017 =\mathrm V = 1.7 \ \mathrm {mV} \$.


How good is your meter?

Meanwhile, what's happening in your battery?

\$ P = I^2R = 20^2 \times 0.6 = 240 \ \mathrm W \$. This may be dangerous.