How to show continuity of a function on a closed interval with continuitiy method and $\epsilon-\delta$ method? Let $I$ be a closed interval and let $f:I\rightarrow I$ satisfy
$ |f(x) - f(y)| \lneq \alpha|x - y| $ for $\alpha > 0$
Show that $f$ is continuous on $I$ by using
i) the definition of continuity method
ii) $\delta$-$\epsilon$ method
I understand that using the continuity method, you have to show that the limit of the function is the same on the left and right side. However, I don't know how to use this to show the continuity of the function.

Let $(x_n)_n$ be an arbitrary sequence with $x_n \to x$. Then
$$0\leq|f(x_n)-f(x)| < \alpha |x_n -x|.$$ 
The right hand goes to $0$ for $n\to +\infty$ what shows that $f$ is continuous (in $x$).
