I don't understand the very last part of the chain rule proof. At this time, we must show $$\lim_{x\to a}\frac{\vert\psi(f(x))\vert}{\vert x-a\vert}=0 \enspace\enspace(6)$$
while knowing that $$\lim_{f(x)\to f(a)}\frac{\vert\psi(f(x))\vert}{\vert f(x)-f(a)\vert}=0 \enspace\enspace(5)$$
Applying the definition of the limit to $(5)$, we can say for any given $\epsilon >0$ that $$\vert\psi(f(x))\vert<\epsilon\vert f(x)-f(a)\vert,$$
whenever $\vert f(x)-f(a)\vert<\delta$ for some $\delta >0$. This latter part happens to be true, whenever $\vert x - a\vert<\delta_1$ where $\delta_1 > 0$. I'm omitting a few details, but what follows is the inequality $$\vert \psi (f(x))\vert<\epsilon\vert f(x)-f(a)\vert\leq\epsilon\vert\phi(x)\vert+M\vert x-a\vert \enspace \enspace (7)$$
where $$ \lim_{x\to a}\frac{\vert\phi(x)\vert}{\vert x-a\vert}=0. \enspace \enspace (8)$$
How does $(6)$ follow from this? Dividing $(7)$ by $\vert x - a\vert$ then taking the limit does not work. For we are left with $M$ on the rhs. Maybe for some reason $(8)$ implies the numerator must also tend to zero when $x\to a$. Well, in that case letting $x\to a$ in $(7)$ would imply $\vert \psi(f(x))\vert \to 0$. But this again does not help. We are merely left with an indeterminate form $0/0$ for the limit $(6)$.
For your information, $\phi(x) = f(x)-f(a)-\lambda(x-a)$, where $\lambda$ is the derivative of $f$ at $a$.
(7) is not required.
$$\lim_{x\to a}\frac{\vert\psi(f(x))\vert}{\vert x-a\vert}$$ can be written as
$$\lim_{x\to a}\frac{\vert\psi(f(x))\vert}{\vert f(x)-f(a)\vert} \frac {\vert f(x)-f(a)\vert} {\vert x-a\vert}$$ The first factor tends to $0$ so we only need boundedness of the second factor as $ x \to a$. But $$\frac {\phi (x)|} {|x-a|} \to 0$$ which implies that $$\frac {\vert f(x)-f(a)\vert} {\vert x-a\vert} \to |\lambda|$$ Can you finish?