Then if the constraint concerns the $L^\infty$ norm, that is, $\forall i \in \{x, y, z\} \leq c$, in each individual direction you have a probelm of a starting velocity, a target endpoint, and a maximum acceleration. You easily solve (for each coordinate) for the time taken to get to that target using the maximal acceleration: this is a matter of simply solving a quadratic equation, using maximum acceleration in the "right" direction. Of those three arrival times, take the longest; that direction will indeed use maximal acceleration. Now for the other two coordinates, fix the arrival time at that longest time, and solve for the necessary acceleration to get home at just that instant; it will in each case be less than the maximum allowed.

So the $L^\infty$ case does not have worries about continuity.

If the constraint is n the $L^2$ sense (that is, $a_x^2+a_y^2+a_z^2 \leq c^2$), then the problem becomes more difficult. First, you can always make a homogeneous linear transformation of variables such that you are trying to travel from $(x_0,0,0)$ to the origin, and indeed there is still a rotation about the $X$ axis degree of freedom left, so that the vector $v_0$ can be considered to lie in the $XY$ plane. These transformations leave the $L^2$ norm constraint unchanged, and turn this into a two-dimensional problem.

I believe the optimal solution to this two-dimensional problem will use a constant acceleration, although this is not so easy to prove. The angle of acceleration is dictated by the requirement that both $x$ and $y$ arrive at zero at the same time; there is in general a unique solution. A bit of algebra arrives at the tangent of the correct angle; again the maximum acceleration magnitude is used, and again there is no continuity worry.
Small perturbations about this solution improve one arrival time, at the cost of making the other later, so this is an extremum.

Note that your problem did not say you have to end up **stopped** at the origin. If it did, you would then run up against the "bang-bang" principle.