If$|f(x)-f(y)|le (x-y)^2$, prove that $f$ is constant

Late answer, but here is a slight generalization of this problem.

Let $X,Ysubseteqmathbb{R}$ and suppose $f:Xto Y$ is a function. Suppose $f$ is $alpha$-Holder continuous with $alphainmathbb{R}$ and $alpha>1$. Then, there exists a $Kinmathbb{R}$ such that for all $x,yin X$, we have $|f(x)-f(y)|leq K|x-y|^{alpha}$. Now, consider $$0leq lim_{xto y}bigg|frac{f(x)-f(y)}{x-y}bigg|leq lim_{xto y}K|x-y|^{alpha-1} $$ Since $alpha>1$, then $alpha-1>0$, which implies $limlimits_{xto y}|x-y|^{alpha-1}=0$. By the squeeze theorem, we have $limlimits_{xto y}bigg|frac{f(x)-f(y)}{x-y}bigg|=0$ which implies $f'(x)=0$. Since the only functions whose derivatives are identically zero are the constant functions, then we must have $f(x)=c$ for some $cinmathbb{R}$.

Technically there is a slight mistake.

You took $xin Bbb{R}$ and $yin Bbb{R}$ such that $yneq x $, so you should $yrightarrow x$. Then you arrive $f'(x)=0$ for all $xin Bbb{R}$, which gives $f$ is constant.

Your deduction that $$frac{|f(x)-f(y)|}{|x-y|}le x-y$$ is incorrect because it would lead to $$|f(x)-f(y)|le |x-y|cdot(x-y)ne (x-y)^2$$ To make it work, you may want to deduce that $$frac{|f(x)-f(y)|}{|x-y|}le |x-y|$$

Your solution is otherwise correct.