Consider an infinite $3$D space with a charge density $\rho$ and a resulting electric field $E$.
Imagine $\forall (x,y,z)\in \mathbb{R}^3, \rho(x,y,z) = \rho_0$(a non-zero constant). In this case, we should expect $\forall (x,y,z), E=0$. This is because of symmetry in charge distribution.
However, according to Gauss's law, $\nabla \cdot E = \frac{\rho}{\epsilon_0}$, divergence of electric field is not zero and neither is the electric field $E$.
I'd appreciate if someone could help me understand what's wrong in this argument.
Maxwell's theory (including Gauss's Law) applies to the analysis of "free" energy in the electromagnetic field; associated for example with the existence of electrical potential differences over finite distances in the field, as in the case of most interacting charge distributions studied in physics and engineering.
If all the observable universe were at the same non-zero potential, with a uniform background charge density existing, then there would exist one more form of spatially distributed energy density that would need to be considered outside the scope of Maxwell's equations.
Gravitational energy and the binding energy associated with the constitution of charged elementary particles, are other examples of spatially distributed energy densities (from the classical field theory viewpoint) that are considered outside the scope of Maxwell's theory.
As J.J Thomson noted in editing the third edition of Maxwell's work; the physical energy density at any point in space (even when limited to studying electromagnetic fields alone) cannot be uniquely determined in Maxwell's Theory.