Answer
False.
Work Step by Step
At $x=0$:
$\lim\limits_{x \to 0} x*sin(\frac{1}{x})=\lim\limits_{x \to 0}\frac{sin(\frac{1}{x})}{\frac{1}{x}}=1$
For small angles, sine has the same sign of its angle, which makes the limits from $x->0^+$ and $x->0^-$ equal, so the limit exists.
But since $f(0)=0$, $\lim\limits_{x \to 0} f(x)\ne f(0)$. Therefore, $f(x)$ is not continuous everywhere.
