Suppose $f$ is continuous on $[0,2]$ and that $f(0) = f(2)$. Then $\exists$ $x,$ $y \in [0,2]$ such that $|y - x| = 1$ and that $f(x) = f(y)$.
Let $g(x) = f(x+1) - f(x)$ on $[0,1]$. Then $g$ is continuous on $[0,1]$, and hence $g$ enjoys the intermediate value property! Now notice $g(0) = f(1) - f(0)$ $g(1) = f(2) - f(1)$ Therefore $ g(0)g(1) = -(f(0) - f(1))^2 < 0$ since $f(0) = f(2)$. Therefore, there exists a point $x$ in $[0,1]$ such that $g(x) = 0$ by the intermediate value theorem. Now, if we pick $y = x + 1$, i think the problem is solved.
I would like to ask you guys for feedback. Is this solution correct? Is there a better way to solve this problem?