A common way to prove this is to first show that $|\sin x| \le |x|$.
For fun we use another approach. Note that if $0 < x < \pi/2$, then $0 < \sin x < 1$ and $0< \cos x< 1$. Recall the familiar identity $\sin 2x =2\sin x\cos x.$ It is more convenient to rewrite this as $\sin u=2\sin \frac{u}{2}\cos \frac{u}{2}.$ If $0, we can rewrite this as $\sin \frac{u}{2}=\frac{1}{2}\frac{\sin u}{\cos\frac{u}{2}}.$ But if $u<\pi/2$, then $\cos\frac{u}{2}>\frac{1}{\sqrt{2}}$, and therefore $\sin \frac{u}{2}<\frac{1}{\sqrt{2}}\sin u.\qquad(\ast)$
Let $u=1$. Since $\sin 1<1$, we find by using $(\ast)$ that $\sin \frac{1}{2}<\frac{1}{\sqrt{2}}.\qquad (1)$ Let $u=\frac{1}{2}$. By using $(\ast)$ again, and $(1)$, we find that $\sin \frac{1}{4}<\left(\frac{1}{\sqrt{2}}\right)^2.\qquad (2)$ Let $u=\frac{1}{4}$. By using $(\ast)$ and $(2)$, we find that $\sin\frac{1}{8}<\left(\frac{1}{\sqrt{2}}\right)^3. \qquad (3)$
Continue. In general we have $0<\sin\frac{1}{2^k}<\left(\frac{1}{\sqrt{2}}\right)^k.$ Thus $\lim_{k\to\infty} \sin\frac{1}{2^k}=0.$ For $0, the sine function is an increasing function. It follows that $\lim_{n\to \infty} \sin\frac{1}{n}=0.$