 Solution

To prove the required zero-existence statement directly, let $S=\{x\in [a,b]:f(x)\le 0\}$ and $c=\sup S$. This set is nonempty and bounded. Continuity and the strict signs at the endpoints give a right neighbourhood of $a$ with negative values and a left neighbourhood of $b$ with positive values; consequently $a<c<b$. If $f(c)<0$, continuity would give points of $S$ to the right of $c$, contradicting its upper-bound property. If $f(c)>0$, continuity would give an interval immediately to the left of $c$ containing no points of $S$, contradicting the approximation property of its supremum. Hence $f(c)=0$ for some $c\in (a,b)$, proving this form of the intermediate value theorem.

For the half-interval claim, put $H(x)=g(x+1/2)-g(x)$ on $[0,1/2]$. It is a continuous function andIf either endpoint value is zero, it supplies the answer. Otherwise the signs are opposite, so the intermediate value theorem, applied also to $-H$ if necessary, gives an interior zero. Thus $g(c+1/2)=g(c)$ for some $c\in [0,1/2]$.

For general positive integer $n$, define $H_n(x)=g(x+1/n)-g(x)$ on $[0,(n-1)/n]$. At its grid points,If a sampled value is zero, it is the desired point. If none is zero, not all sampled values can have the same sign, so two have opposite signs; continuity gives a zero between their grid points. This proves the horizontal chord of reciprocal-integer length:For $n=1$ the interval is the single point zero, which works by the endpoint equality itself. The proof does not require a differentiable function or a periodic extension.