Show that $\{f_n\}$ has a uniformly convergent subsequence. 
Let the function $f_n$ : $[0,1] \rightarrow [0,1]$ satisfy
  \begin{align*}
\vert f_n(x)-f_n(y)\vert \leq \vert x-y \vert \textrm{ whenever } \vert x-y \vert \geq \frac{1}{n}
\end{align*} 
  Show that the sequence $\{f_n\}$ has a uniformly convergent subsequence. 

I try to use the Arzela-Ascoli Theorem. Thus, I check to $\{f_n\}$ is uniformly bounded and equicontinuous. But, I don't know how to prove that when $\vert x - y \vert <\frac{1}{n}$
Any help is appreicated...
Thank you!
 A: Warning: this solution assumes that the functions $f_n$ are continuous on $[0,1]$. In general Arzelà-Ascoli is not applicable here.
Let $\varepsilon > 0$. Pick $n \in \mathbb{N}$ such that $\frac5n < \varepsilon$. For $|y-x| < \frac1{n}$ and $k \ge n$
$$\left|\left(x-\frac2n\right) - x\right| = \frac2n \ge \frac1n \ge \frac1k$$
$$\left|y-\left(x-\frac2n\right)\right| \ge \frac2n-  |y-x| \ge \frac1{n} \ge \frac1k$$
so
\begin{align}
|f_k(x) - f_k(y)| &\le \left|f_k(x) - f_k\left(x-\frac2n\right)\right| + \left|f_k\left(x-\frac2n\right) - f_k(y)\right|\\
&\le \left|\left(x-\frac2n\right) - x\right| + \left|y-\left(x-\frac2n\right)\right| \\
&\le \frac2n + |y-x| + \frac2n\\
&< \frac5{n}\\
&< \varepsilon
\end{align}
Each $f_j$ is uniformly continuous on $[0,1]$ so there exists $\delta_j > 0$ such that $|y-x| < \delta_j$ implies $|f_j(x) - f_j(y)| < \varepsilon$.
Therefore for $|y-x| < \delta =\min\left\{\delta_1, \ldots, \delta_{n-1}, \frac1n\right\}$ we have $|f_j(x) - f_j(y)| < \varepsilon, \forall j \in \mathbb{N}$.
Hence $(f_n)_n$ is equicontinuous. $(f_n)_n$ is also uniformly bounded by $1$. 
By Arzelà-Ascoli, there exists a uniformly convergent subsequence of $(f_n)_n$.
A: Although functions $f_n$ are not equi-continuous (and not even continuous), one can still apply the Arzela-Ascoli diagonalization argument, with a slight preparation.
Let $n\in \mathbb{N}$ be fixed. Take any $x_1\in (0,1)$ and consider the set $$U_{x_1} = (x_1 - \frac 2n, x_1 + \frac 2n) \setminus [x_1-\frac 1n,x_1 + \frac 1n].$$ Then $U_{x_1}$ is an open set, and for any $x\in U_{x_1}$ we have, by the condition of the problem, that
$$
| f_n(x)  - f_n(x_1) |\leq \frac 2n.
$$
Hence, for any $x,y \in [0,1]$ with $|x-x_1|, |y-x_1| \leq 1/n$ we get
$$
\tag{1} |f_n(x) - f_n(y)| \leq |f_n(x) - f_n(x_1) | + |f_n(y) - f_n(x_1)| \leq \frac 4n.
$$
Let $x_1$ run over rationals $Q$ (say), then $\{(x_1 - 1/n, x_1 + 1/n)\}_{x_1 \in Q}$ will serve as an open cover of $[0,1]$. Extracting a finite sub-cover and applying $(1)$ to pass from one interval of length $2/n$ to its neighbor, we get
$$
\tag{2} |f_n(x) - f_n(y) |\leq \max\left\{ \frac 8n, |x-y| \right\}.
$$
Now, $(2)$ shows "asymptotic equi-continuity", the oscillations of $f_n$ decay uniformly as $n\to \infty$, and we are now ready to apply the argument in the proof of Arzela-Ascoli using $(2)$ instead of the usual equi-continuity.
