Solution of A Horizontal Chord of Half the Length of the Interval

Let $J=[0,\tfrac12]$, which is a closed interval with $0<\tfrac12$. Define $g:J\to\mathbb{R}$ by

$$g(x)=f\!\left(x+\tfrac12\right)+(-1)\,f(x) .$$

This is well defined: if $x\in J$ then $0\le x\le\tfrac12$, so $\tfrac12\le x+\tfrac12\le 1$ and hence $x+\tfrac12\in[0,1]$.

Step 1: $g$ is continuous on $J$. Let $h:J\to\mathbb{R}$ be given by $h(x)=f(x+\tfrac12)$. Fix $x\in J$ and let $\varepsilon>0$. Since $x+\tfrac12\in[0,1]$ and $f$ is continuous at $x+\tfrac12$, there is $\delta>0$ such that every $w\in[0,1]$ with $|w-(x+\tfrac12)|<\delta$ satisfies $|f(w)-f(x+\tfrac12)|<\varepsilon$. Now let $z\in J$ with $|z-x|<\delta$, and put $w=z+\tfrac12$. As above $w\in[0,1]$, and

$$\left|w-\left(x+\tfrac12\right)\right|=|z-x|<\delta ,$$

so $|h(z)-h(x)|=|f(w)-f(x+\tfrac12)|<\varepsilon$. Hence $h$ is continuous at $x$ relative to $J$, and since $x\in J$ was arbitrary, $h$ is continuous on $J$.

The restriction of $f$ to $J$ is continuous on $J$ by clause 1 of Restriction Stability of Continuity and of the Derivative, since $J\subseteq[0,1]$. As $g$ is the sum of $h$ and $(-1)$ times that restriction, $g$ is continuous on $J$ by clauses 2, 4 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space.

Step 2: the endpoint values of $g$ are negatives of one another. Put $t=g(0)=f(\tfrac12)-f(0)$. Using the hypothesis $f(1)=f(0)$,

$$g\!\left(\tfrac12\right)=f(1)-f\!\left(\tfrac12\right)=f(0)-f\!\left(\tfrac12\right)=-t .$$

Step 3: applying the intermediate value theorem. The order of $\mathbb{R}$ is total, so $t\le 0$ or $0\le t$. By Elementary Order Arithmetic in an Ordered Field, $t\le 0$ implies $0\le -t$, and $0\le t$ implies $-t\le 0$. Hence:

if $t\le 0$, then $g(0)=t\le 0\le -t=g(\tfrac12)$;

if $0\le t$, then $g(\tfrac12)=-t\le 0\le t=g(0)$.

In either case the value $0$ satisfies one of the two alternatives in the hypothesis of Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals for $g$ on $[0,\tfrac12]$. Since $0<\tfrac12$ and $g$ is continuous on $[0,\tfrac12]$, that theorem gives $c\in[0,\tfrac12]$ with $g(c)=0$, that is,

$$f\!\left(c+\tfrac12\right)=f(c) ,$$

which is what was required.