Let $f\colon\mathbb{Q}\to\mathbb{R}$ be a function defined on the set $\mathbb{Q}^n\subset\mathbb{R}^n$ consisting of points whose coordinates are rational numbers. Prove that if $f$ satisfies the inequality $|f(x)-f(y)|^{2018}\leq2018|x-y|$ for all $x,y\in\mathbb{Q}^n$, then there is a unique continuous function $F\colon\mathbb{R}^n\to\mathbb{R}$ such that $F(x)=f(x)$ for all $x\in\mathbb{Q}^n$. Provide a direct proof without referring to any deep results.
I will be glad if someone will proof-read my solution, since I'm not sure how "deep" are the results that should be avoided.
First of all, note that $$|f(x)-f(y)|^{2018}\leq2018|x-y|$$ implies $$|f(x)-f(y)|\leq\sqrt[2018]{2018}|x-y|^{\frac{1}{2018}},$$ so $f$ is $\alpha$-Holder continuous with $\alpha=\frac{1}{2018}$, therefore uniformly continuous on $\mathbb{Q^n}$, which is dense in $\mathbb{R}^n$.
Define $$F(x)=\lim\limits_{x_k\to x}f(x_k)$$ for all $x\in\mathbb{R}^n\setminus\mathbb{Q}^n$, where $x_k\in\mathbb{Q}^n$. Need to show that such $F$ is unique and continuous.
Uniqueness follows from uniqueness of the limit. Uniformly continuous $f$ is Cauchy-continuous, i.e., maps Cauchy sequences to Cauchy sequences, thus extends to continuous function on the closure of $\mathbb{Q}^n$, which is $\mathbb{R}^n$.
Can my explanation of continuity of $F$ be indeed considered "elementary" or the referred result is pretty "deep"? Does this problem admit more elementary solution?
To show that $F$ is well defined you have to show that if $x_k \to x$ and $y_k \to x$ ($x_k,y_k \in \mathbb Q^{n}$) then $\lim f(x_k)=\lim f(y_k)$. For this use the fact that the sequence $(z_k)$ defined by $z_{2k}=x_k$ and $z_{2k-1} =y_k$ also converges to $x$ so $\lim f(z_k)$ exists. This implies that $\lim f(x_k)=\lim f(y_k)$. For continuity of $F$ simply take limits in the inequality that gives Holder continuity of $f$ and you will see that $F$ is in fact Holder continuous.