Prove: The following $[x]+2$ numbers
$$1, \quad j \alpha-[j \alpha], \quad j=0,1, \cdots,[x]$$

are all in the interval $[0,1]$, and thus by the pigeonhole principle, there must be two numbers whose difference does not exceed $([x]+1)^{-1}$. If these two numbers are
$$j_{1} \alpha-\left[j_{1} \alpha\right], \quad j_{2} \alpha-\left[j_{2} \alpha\right], \quad 0 \leqslant j_{1}<j_{2} \leqslant[x],$$

then we take $d=j_{2}-j_{1}, c=\left[j_{2} \alpha\right]-\left[j_{1} \alpha\right]$ and
$$a=c /(c, d), \quad b=d /(c, d) .$$

Otherwise, these two numbers must be
$$1, \quad j_{1} \alpha-\left[j_{1} \alpha\right], \quad 0 \leqslant j_{1} \leqslant[x]$$

In this case, we take $d=j_{1}, c=\left[j_{1} \alpha\right]+1$ and
$$a=c /(c, d), \quad b=d /(c, d) .$$

It is easy to verify that, in either case, the chosen $a, b$ satisfy equation (10), and
$$|\alpha-a / b| \leqslant 1 /(b([x]+1)),$$

from which it follows that equation (11) holds. Proof complete.