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.

Fix $x_0 \in I$. Then we have
$ |f(x) - f(x_0)| < \alpha|x - x_0| $ .
For $x \to x_0$ we have $\alpha|x - x_0|  \to 0$ , thus $ |f(x) - f(x_0)| \to 0$, hence
$f(x) \to f(x_0)$
