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 $0\frac{1}{\sqrt{2}}$, 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