Help to understand proof to show that a function is uniformly continous in a certain interval (Spivak)

Help to understand proof to show that a function is uniformly continous in
a certain interval (Spivak)

This excerpt relates to the following proof:
If $f$ is continous on $[a,b]$, then $f$ is uniformly continuous $[a,b]$.
For $\epsilon > 0$ let's say that $f$ is $\epsilon$-good on $[a,b]$ if
there is some $\delta > 0$ such that, for all $y$ and $z$ in $[a,b]$,
if $|y-z|<\delta$ , then $|f(x)-f(z)|<\epsilon$
Consider any $\epsilon > 0$. Let
$A=\{x:a\leq x \leq b$ and $f$ is $\epsilon$-good on $[a,x]\}$
Then, the proof went on to show that $A$ has a least upper bound, $\alpha$
and that $\alpha=b$. Here's the part I don't understand:
... So $f$ is surely $\epsilon$-good on the interval $[\alpha - \delta_0,
\alpha + \delta_0]$. On the other hand, since $\alpha$ is the least upper
bound of $A$, it is also clear that $f$ is $\epsilon$-good on $[a, \alpha-
\delta_0]$.
I don't understand how we can conclude that $f$ is $\epsilon$-good $[a,
\alpha- \delta_0]$ just from knowing that $\alpha$ is the least upper
bound of $A$. How do we know that $f$ is $\epsilon$-good for each number
in the interval from $a$ up to and including $\alpha-\delta_0$?
The proof also claims that $f$ is $\epsilon$-good on $[a, b - \delta_0]$
which I also don't understand but I guess the explanation would be similar
to $f$ being $\epsilon$-good on $[a,\alpha-\delta_0]$.
Thank you in advance for any help provided.