To test whether $|0\rangle$ is physical, you would apply $\partial^\mu A_\mu^+$ to it. So $|0\rangle \in V$.

To test whether $|0\rangle$ is physical, you would apply $\partial^\mu A_\mu^+$ to it. So $|0\rangle \in V$.

Note that 7.48 does not mean that $\partial_\mu A^\mu = 0$ as an operator identity in this space. It means that all it's matrix elements in this space are zero. You have noticed that the state you created, $\partial_\mu A^\mu |0\rangle$, lives not in $V$ but in the space orthogonal to it. So it has zero overlap with any states in $V$, so all the matrix elements of $\partial_\mu A^\mu$ are still zero in $V$.

Long story short, we can't impose $\partial_\mu A^\mu = 0$ as an operator identity without losing the canonical commutation relations.