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 $$\lim_{n\to \infty} \sin\frac{1}{n}=0.$$