x times the Dirichlet function

At $0$. For every $x$, $0\le|g(x)|=|x|\,D(x)\le|x|$. By the squeeze theorem, $g(x)\to0$ as $x\to0$. Since $g(0)=0$, the function $g$ is continuous at $0$.

At $a\ne0$. For every $n\ge1$, the interval $\left(a,a+\frac1n\right)$ contains a rational number $r_n$ and an irrational number $s_n$. We thus have $r_n\to a$, $s_n\to a$ and
$$g(r_n)=r_n\longrightarrow a,\qquad g(s_n)=0\longrightarrow0.$$
Since $a\ne0$, these two limits are different, so $g$ has no limit at $a$.

Remark. The function $g$ is continuous at exactly one point, $0$. Continuity at a single point is possible, and it comes from the factor $x$ squeezing $D$.