For part a, as you mentioned,
we can use the...
...which says that, as is continuous in x and Lipschitz continuous in y, there is a unique solution to the initial value problem, which is defined in a neighbourhood of 0.
Now for part b, note that is a solution of the equation. If was to meet somewhere over , the uniqueness of the initial value problem would be violated. Since , we must have throughout . Now the ode gives us a constant sign for over .