This is not true: take $\vert \cdot \vert :\mathbb{R} \rightarrow \mathbb{R}$ (absolute value) is Lipschitz but $x^2$ is not. If $D$ is bounded however use $\vert f(x)g(x) - f(y)g(y) \vert \leq \vert f(x) \vert \vert g(x)-g(y) \vert + \vert g(y) \vert \vert f(x)-f(y) \vert$