Simplify $(\nabla \times \mathbf{B})\times \mathbf{B}$ for $\nabla \cdot \mathbf{B} = 0$.
Using index notation: $$[(\nabla \times \mathbf{B})\times \mathbf{B}]_i = \epsilon_{ijk}\epsilon_{jlm} \partial_lB_mB_k = \epsilon_{kij}\epsilon_{jlm}\partial_lB_mB_k = (\delta_{kl}\delta_{im}-\delta_{km}\delta_{il})\partial_lB_mB_k = \partial_kB_iB_k-\partial_iB_kB_k = B_k \partial_k B_i + \underbrace{B_i\partial_kB_k}_\text{=0}-\partial_iB_kB_k = ... = (\mathbf{B} \cdot \nabla)\mathbf{B} - \nabla(\mathbf{B} \cdot\mathbf{B})$$
However, my book insists that $\partial_i B_k B_k = [\nabla(\dfrac{1}{2} \mathbf{B} \cdot\mathbf{B} )]_i$. What is correct?
You need to distinguish between the derivative operating on one part of the product and operating on all of it. The conventional way to do this is to be quite generous with the brackets: $ (\partial_i B_k)B_k \neq \partial_i (B_kB_k) $. Since the $\nabla \times$ acts only on the first $\mathbf{B}$, all the derivatives in the initial expansion of the product only act on one $B_i$, i.e. you have \begin{align} [(\nabla \times \mathbf{B}) \times \mathbf{B}]_i &= \epsilon_{ijk}\epsilon_{jlm} (\partial_l B_m) B_k \\ &= (\delta_{kl}\delta_{im}-\delta_{km}\delta_{il}) (\partial_l B_m) B_k \\ &= (\partial_k B_i) B_k - (\partial_i B_k)B_k \\ &= (\partial_k B_i) B_k - \partial_i (\tfrac{1}{2}B_k B_k) \\ &= \left[(\mathbf{B} \cdot \nabla )\mathbf{B}- \nabla\tfrac{1}{2} \lvert \mathbf{B} \rvert^2 \right]_i \end{align}
Intuitively, the condition $\nabla \cdot \mathbf{B}=0$ doesn't come into it: such a term would have to look like $(\nabla \cdot \mathbf{B})\mathbf{B}$, which is parallel to $\mathbf{B}$ and could not be part of the expansion of a cross product containing $\mathbf{B}$.