A continuous function that vanishes on the rationals

1. Let $x\in\mathbb{R}$. Take $q_n=\dfrac{\lfloor 10^nx\rfloor}{10^n}$, a rational number with $x-10^{-n}<q_n\le x$, so $q_n\to x$. Since $f$ is continuous at $x$, $f(q_n)\to f(x)$. But $f(q_n)=0$ for all $n$, so $f(x)=0$.

2. Apply part 1 to the continuous function $f-g$, which vanishes on $\mathbb{Q}$.

Remark. A continuous function is entirely determined by its values on a dense set. This is the principle used to define $a^x$ for real $x$: we know $a^q$ for rational $q$, and there is only one continuous way to fill the gaps. Without continuity the statement is false: the Dirichlet-type function $\mathbf 1_{\mathbb{R}\setminus\mathbb{Q}}$ vanishes on $\mathbb{Q}$ without being $0$.