Suppose $f(x)=xg(x)$, where $g$ is a continuous at $x_0=0$. Then $f$ is differentiable at $x_0=0$. 
Justify: Suppose $f(x)=xg(x)$, where $g$ is a continuous at $x_0=0$. Then $f$ is differentiable at $x_0=0$.

I tried proving this by contradiction, but I'm not sure this is correct.
My attempt: Supposed $f$ is not differentiable at $x_0=0$. Then the following limit must not exist.
$$\begin{align}\lim_{x\to0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to0} \frac{xg(x)-0}{x-0} &= \lim_{x\to0} g(x)= g(0).\end{align}$$
That means $f'(0)=g(0)$, which is a contradiction.
Is this correct? If so, are there other ways to prove this?

First you should show that $f(0) = 0$ (even if it is simple).
Then you have the right idea, but make it unnecessary complicated by trying to wrap it in a proof by contradiction. It suffices to show that
$$
\lim_{x\to0} \frac{f(x)-f(0)}{x-0} 
$$
exists, because that means that $f$ is differentiable at $x_0 = 0$.
