If f is continuous for every real number and $f(r)=0$ for every rational number, then $f(x)=0$ for all real numbers. Am I going in the right direction? 
Suppose that $f: \mathbb{R} \to \mathbb{R}$ is continuous on $\mathbb{R} $ and that $f(r)=0$ for every rational number $r$. Prove that $f(x)=0$ for all $x$ in $\mathbb{R}$.

Proof: 
Since $r \in \mathbb R $, then $\forall \epsilon \gt 0$, $\exists \delta \gt 0$ such that if $\lvert x-r \rvert \lt \delta$ $\forall x \in \mathbb R $, then $\lvert f(x)-f(r) \rvert = \lvert f(x) \rvert \lt \epsilon$.
Since $\lvert f(x) \rvert \lt \epsilon$, it follows that $f(x)=0$ $\forall x \in \mathbb R $.

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, for every real number $x \in \mathbb{R}$, there exists a sequence $(q_n)_{n=1}^\infty \in \mathbb{Q}$ such that $q_n \to x$ (for example, take the sequence of truncated decimal expansions). Then, by continuity,
$$f(x) = f\left(\lim_{n \to \infty} q_n\right) = \lim_{n \to \infty}f(q_n) = \lim_{n \to \infty} 0 = 0.$$
