I believe @Phillip Wood is correct in his comment about the filling of the space with the dielectric, and not just partially filling it. It is when the space is filled that the polarization of the dielectric reduces the net electric field within the filled space. In other words, in your example the reduction is only within your little dipole.
Regarding the conclusions you have reached about your simulation, perhaps you have been following the comments between @Phillip Wood and I. Your simulation shows that the potential at the location of the test charge appears to increase because of the dipole. This is true but the degree to which it increases depends on the ratio of the distance between the test charge and the dipole to the separation distance between the charges of the dipole. The greater that ratio, the more the dipole looks like a net charge of zero, and the less the increase in potential.
So unless you have already done so, you need to factor into your simulation the ratio of the distance between the test charge and the dipole to the distance between the charges in the dipole. The greater that ratio is, the less the influence of the dipole on the test charge.
In closing, I want to thank Philip Wood for his insight.
Hope this helps.