I'll show that $f(0)=f(1)$ and you'll see the trick. For all $n\in \mathbb{N}$, $n > 0$:

$$ |f(1)-f(0)| = |f(1)-f(\frac{n-1}{n})+f(\frac{n-1}{n})-f(\frac{n-2}{n}) + \dotsc + f(\frac{1}{n}) - f(0)| \leq \\ \frac{1}{n^2} + \frac{1}{n^2} + \dotsc + \frac{1}{n^2} = \frac{1}{n} $$

(There are $n$ terms in that sum.) Since this holds for all positive $n$ it follows that $f(1) = f(0)$.

I'll show that $f(0)=f(1)$ and you'll see the trick. For all $n\in \mathbb{N}$, $n > 0$:

\begin{align} |f(1)-f(0)| &= |f(1)-f(\frac{n-1}{n})+f(\frac{n-1}{n})-f(\frac{n-2}{n}) + \dotsc + f(\frac{1}{n}) - f(0)| \newline &\leq \frac{1}{n^2} + \frac{1}{n^2} + \dotsc + \frac{1}{n^2} = \frac{1}{n} \end{align}

(There are $n$ terms in that sum.) Since this holds for all positive $n$ it follows that $f(1) = f(0)$.