In Spivak's book there is a theorem proved that if $f$ is continuous on $[a,b]$ then $f$ is uniformly continuous on $[a,b]$ which starts as follows:
For $\epsilon>0$ let the function be $\epsilon$-good on $[a,b]$ if there is some $\delta>0$ such that for all $y$ and $z$ $\in [a,b]$:
If $\lvert y-z\rvert<\delta$, then $\lvert f(y) - f(z)\rvert<\epsilon$
Consider any particular $\epsilon>0$.
Let $A = \bigl[\space x:a\le x\le b$ and $f$ is $\epsilon$-good on $[a,x]\space\bigr]$
Clearly $a\in A$ and $A$ is bounded above (by b), so the least upper bound $\alpha$ exists.
If $\alpha<b$, then because $f$ is continuous, there is $\delta_0>0$ such that:
$\lvert y-\alpha\rvert<\delta_0 \implies \lvert f(y)-f(\alpha)\rvert<\frac\epsilon2$
Consequently if $\lvert y-\alpha\rvert<\delta_0$ and $\lvert z-\alpha\rvert<\delta_0$ then $\lvert f(y)-f(z)\rvert<\epsilon$.
So $f$ is surely $\epsilon$-good on $[\alpha-\delta_0,\space\alpha+\delta_0]$
I am unable to reproduce the last conclusion.
By definition, in order for $f$ to be $\epsilon$-good on $[\alpha-\delta_0,\space\alpha+\delta_0]$ for $\epsilon>0$, we must find $\delta>0$ such that for all $y,z\in[\alpha-\delta_0,\space\alpha+\delta_0]$:
If $\lvert y-z\rvert<\delta$, then $\lvert f(y) - f(z)\rvert<\epsilon$
I suspect that in order to do so, we want for some $\delta$: $\lvert y-z\rvert<\delta \implies \lvert y-\alpha\rvert<\delta_0$ and $\lvert z-\alpha\rvert<\delta_0$, however I am definitely missing some obvious step that would allow me to reach this conclusion.
If $x,y\in[\alpha-\delta_0,\alpha+\delta_0]$, then $|f(x)-f(\alpha)|<\frac\varepsilon2$ and $|f(y)-f(\alpha)|<\frac\varepsilon2$. Therefore,\begin{align}|f(x)-f(y)|&=\left|\bigl(f(x)-f(\alpha)\bigr)-\bigl(f(y)-f(\alpha)\bigr)\right|\\&\leqslant|f(x)-f(\alpha)|+|f(y)-f(\alpha)\\&<\varepsilon.\end{align}