You need that $f(0)=0$ otherwise you could take $f(x)=1$ which is continuous at $x_0=0$ but you have that $f(x)\cdot D(x)=D(x)$ which is not continuous at $0$.
If you have it however it is easy to see that is correct by using the fact that $|D(x)|\leq1$ everywhere. Therefore you have that $f(x)D(x)$ has to be $0$ at $x_0=0$ (I let you write out the details).
Edit: As you are asking for more details, I will give you another hint but the question is very trivial so I fear to completely solve it...
For $g(x)$ to be continuous at $0$ it is sufficient to show that
$\lim_{x \rightarrow 0}g(x)=g \Big(\lim_{x \rightarrow 0}x \Big)=g(0)=f(0)D(0)=0.$
Therefore you can use that
$\lim_{x \rightarrow 0}|g(x)|=\lim_{x \rightarrow 0}|f(x)D(x)|\leq \cdots \leq 0.$
Then your theorem follows, now you have to fill out the dots.