This question is based on this answer.
I don't understand why the following must be true:
The ϵ-goodness of $f$ on $[a,y]$ implies its ϵ-goodness on any subinterval of $[a,y]$.
This is my attempt at proving why it must be true:
$f$ is $\epsilon$-good on $[a,y]$ means that given the $\epsilon>0$, there is a $\delta>0$ such that for all $c,d \in [a,y]$:
If $|c-d|<\delta$, then $|f(c)-f(d)|<\epsilon$.
Therefore, the following must be true: $|a-y| < \delta$ and $|f(a)-f(y)|<\epsilon$
So, if we pick any number, $x$ such that $a<x<y$, then:
$|a-x|<\delta$ is true.
From here, I don't know how to conclude that $|f(a)-f(x)|<\epsilon$ will be true too.
Your proof is wrong. If $f$ is $\epsilon$-good on $[a,y]$, then there is a $\delta >0$ such that for any $c,d\in [a,y]$, if $|c-d|<\delta$ then $|f(c)-f(d)|<\epsilon$. Let $[p,q]$ be a subinterval of $[a,y]$. Then for any $c,d\in [p,q]$, we have $c,d\in [a,y]$, so if $|c-d|<\delta$ then $|f(c)-f(d)|<\epsilon$. This means $f$ is $\epsilon$-good on $[p,q]$.