In QAOA, why do we pick the initial Hamiltonian $B$ to be $sigma_x$ applied to each qubit?

We don't really need $B = sum sigma_j^x$ in our QAOA algorithm. As long as you pick it in such a way that it doesn't commute with $C$. One of the reason is if they are commute, then they share a common eigenvector. Then if you run into this type of situation, you will never get out, and you will be stuck in this state. You can think of $U(beta, B)$ as a driver, it helps to navigate the Ansatze from getting stuck.

In term of the reason why $B = sum sigma_j^x$ in the first place is because QAOA is sort of a discretization of Quantum Annealing so that is why we see that the Ansatze of QAOA takes the form: $U = e^{-ibeta_p B}e^{-i gamma_p C} cdots e^{-ibeta_1 B} e^{-i gamma_1 C} = prod_{i} e^{-ibeta_i B} e^{-i gamma_i C} $ which is a trotter approximation of the time evolution in the quantum annealing.