First, note that the zeroes of $\sin x$ are precisely the integer multiples of $\pi$.
By Hurwitz' Theorem), there are infinitely many pairs of relatively prime integers $a, b$ such that $$\left\vert\pi - \frac{a}{b}\right\vert < \frac{1}{\sqrt{5} b^2},$$ and in particular we can approximate $\pi$ arbitrarily closely with such $(a, b)$. Multiplying through by $b$ gives that $$|b \pi - a|< \frac{1}{\sqrt{5}b}.$$
Since the derivative of $\sin x$ is bounded in magnitude by $1$, we have for each pair $(a, b)$ that $$|\sin a| = |\sin(b \pi - a)| \leq |b \pi - a| < \frac{1}{\sqrt{5} b}.$$
The right-hand side can be made arbitrarily small by taking large enough $b$, and so we can find (corresponding) $a$ such that $|\sin a|$ is as close to zero as desired.