Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

(N/A) Consider a closed equipotential surface $S$ that encloses a volume $V$ containing no charge $(q_{in} = 0)$.
Assume, for the sake of contradiction, that the potential inside the volume is not constant. If the potential varies, there must exist a non-zero electric field $\vec{E}$ within the volume, given by the relation $\vec{E} = -\nabla V$.
According to Gauss's Law, the flux of the electric field through any closed surface is proportional to the enclosed charge: $\oint_S \vec{E} \cdot d\vec{A} = \frac{q_{in}}{\epsilon_0}$.
Since $q_{in} = 0$, the net flux through the surface must be zero.
If the potential were to vary inside, the electric field lines would have to originate or terminate on charges within the volume. However, since there is no charge inside, any field line entering the volume must also exit it, or the field must be zero everywhere.
If the potential were not constant, there would be local maxima or minima of potential inside the region. According to the properties of harmonic functions (Laplace's equation $\nabla^2 V = 0$), the potential cannot have a local maximum or minimum in a charge-free region.
Therefore, the potential must be constant throughout the volume, meaning the entire volume is equipotential.