Let $ABC$ a triangle. Let $T$ a point in the circumcircle such that $BT$ is a diameter.

So $\angle A=\angle CAB$ is equal to $\angle CTB$ (they subtend the same arc). Since $\triangle CBT$ is a right triangle, from the definition of sine we get $$\sin \angle CTB =\frac{BC}{BT}=\frac{a}{2R}.$$

On the other hand $\angle CAB=\angle CTB$ implies their sines are the same and so $$\sin \angle CAB=\frac{a}{2R}$$ and therefore $$\frac{a}{\sin A}=2R.$$

Drawing diameters passing by $C$ and $A$ will let us prove in a similar way the relations $$\frac{b}{\sin B}=2R\quad\mbox{and}\quad\frac{c}{\sin C}=2R$$ and we conclude that $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R.$$

Q.E.D.