Is "The curl of the gradient of any scalar field is identically zero" contradictory to Faraday's Law?

The theorem is about fields, not about physics, of course. The fact that dB/dt induces a curl in E does not mean that there is an underlying scalar field V which corresponds to that E-field. Only conservative electric fields have a representation as gradient of the scalar potential.

In the presence of a changing B field, E is not conservative, and V is undefined (well, at least poorly defined, and not easy to measure).

The correct result is $\mathbf{E}=-\boldsymbol{\nabla}V-\frac{\partial \mathbf{A}}{\partial t}$ with $\mathbf{B}=\boldsymbol{\nabla}\times \mathbf{A}$.

$\vec{E} = -\vec\nabla V$ is true only in electrostatics, in general you cannot write the electric field as a gradient of some scalar function.

Faraday's law is more fundamental than $\vec{E} = -\vec\nabla V$. Faraday's law is always correct, $\vec{E} = -\vec\nabla V$ is not.