Dirichlet's approximation theorem (Farey dissection)

Consider the $Q+2$ numbers $0,\{\alpha\},\{2\alpha\},\dots,\{Q\alpha\},1$, where $\{x\}$ denotes the fractional part of $x$; all lie in $[0,1]$. Partition $[0,1]$ into $Q+1$ equal subintervals of length $1/(Q+1)$: $[0,\tfrac1{Q+1}), [\tfrac1{Q+1},\tfrac2{Q+1}),\dots$.

We have $Q+2$ numbers and only $Q+1$ subintervals, so by the pigeonhole principle two of the numbers $\{j\alpha\}$ and $\{i\alpha\}$ (with $0\le i<j\le Q$, allowing $i=0$ so $\{i\alpha\}=0$) land in the same subinterval, hence differ by less than $1/(Q+1)$: $|\{j\alpha\}-\{i\alpha\}|<\tfrac{1}{Q+1}$.

Set $q=j-i$, so $1\le q\le Q$. Since $\{j\alpha\}-\{i\alpha\} = (j\alpha - i\alpha) - (\lfloor j\alpha\rfloor - \lfloor i\alpha\rfloor) = q\alpha - a$ for the integer $a=\lfloor j\alpha\rfloor-\lfloor i\alpha\rfloor$, we get $|q\alpha - a| < \tfrac{1}{Q+1}$, i.e. $\left|\alpha-\dfrac aq\right| < \dfrac{1}{q(Q+1)}$.

Finally, if $\gcd(a,q)=d>1$, divide both $a$ and $q$ by $d$: the resulting fraction has an even smaller denominator and the same (or a smaller) distance to $\alpha$, so we may take $\gcd(a,q)=1$ without loss of generality. This completes the proof; it is a finite, constructive pigeonhole argument, with no appeal to any unproved estimate.