= Solution
{c}
For $n=1$, choose $x=0$, $y=1$. For $n\geq2$, define the <continuous function>
$$
g(t)=f(t+1/n)-f(t),\qquad 0\leq t\leq1-1/n.
$$
Its values at the grid points have the <telescoping sum>
$$
\sum_{k=0}^{n-1}g(k/n)=\sum_{k=0}^{n-1}\bigl(f((k+1)/n)-f(k/n)\bigr)=f(1)-f(0)=0.
$$
If one summand is zero, we already have a zero of $g$. Otherwise the summands cannot all have the same sign, so two of these values have opposite signs. The <intermediate value theorem> between their arguments gives $t\in[0,1-1/n]$ with $g(t)=0$. Set $x=t$, $y=t+1/n$. Then
$$
\boxed{|x-y|=1/n,\qquad f(x)=f(y).}
$$
This is the <horizontal chord of reciprocal-integer length> property; the <telescoping sum> explains why reciprocal <integers> appear.
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