 Solution

If $D\ge 1/2$, both nonconstant coefficients of $\Delta (q)$ are nonnegative, so it is positive for $q\ge 0$. If $D_c<D<1/2$, its discriminant is negative, so it is again strictly positive. Therefore $\boxed{D>D_c⟹\text{no diffusion-driven instability}}$.

At $D=D_c$ the double root can only give a neutral zero eigenvalue if it coincides with an allowed spatial mode; it still gives no positive growth rate. This is the finite-wavenumber onset of a Turing instability, not an oscillatory crossing since the mode trace stays negative.