Why is the divergence of induced electric field zero?

I think you're drawing a distinction that doesn't need to be drawn. There are two equations that the electric field must satisfy, Gauss's Law & Faraday's Law: $$ vec{nabla} cdot vec{E} = fracrho{epsilon_0}, qquad vec{nabla} times vec{E} = - frac{partial vec{B}}{partial t}. $$ The total electric field always satisfies these equations; that's what Maxwell's equations, taken as a whole, require. So in any situation (even when fields are changing), it is always the case that the divergence of $vec{E}$ is proportional to $rho$. And in any situation (even when there are charges), it is always the case that the curl of $vec{E}$ is equal to the negative of the rate of change of $vec{B}$.

If it happens to be the case that everything is static, then it follows that $vec{E}$ is curl-free; this is the domain of electrostatics. And if it happens to be the case that there are no charges in a particular region, then $vec{E}$ is divergence-free; this is the case of "induced fields" that you're describing above. But it's better, in general, to think about the electric field as a whole rather than as the superposition of some "static field" and some "induced field".