 Solution

For a chord of length $1/2$, consider the continuous function $h(x)=f(x+1/2)-f(x)$ on $[0,1/2]$. Since $f(0)=f(1)$,An endpoint zero already supplies the chord. Otherwise the endpoint values have opposite signs, so the intermediate value theorem supplies $h(\alpha )=0$ at some $\alpha \in (0,1/2)$. Thus a horizontal chord of length $1/2$ exists.

For each integer $n>1$, define $h_n(x)=f(x+1/n)-f(x)$ on $[0,1-1/n]$. The sampled differences telescope:If any sampled value is zero, it gives the required chord. Otherwise there must be both a positive and a negative sampled value. The intermediate value theorem, applied between their sample points, then gives $h_n(\alpha )=0$. Hence a horizontal chord of length $1/n$ exists for every positive integer $n$; for $n=1$, the endpoint chord itself suffices. This proves the horizontal chord of reciprocal-integer length result, without restricting $n$ to powers of two.