For the following short proof, I would like to know if I correctly identified the need to assume that $\phi$ must be continuous on some interval around its limit point and for any interval around its limit point, $\phi$ cannot be constant?
Consider the function $\phi$ that is continuous on some interval around $b$ and, for any interval around $b$, $\phi$ is non-constant. Then, if $\lim_{y \to b}\phi(y)=a$ and if for some function $F$, we have $\lim_{y \to b}F \circ \phi (y)=L$, then $\lim_{x \to a}F(x)=L$.
By assumption, we know that for an arbitrary $\varepsilon \gt 0$, there is an $\eta \gt 0$ such that any $y\in [b-\eta,b+\eta]\setminus\{b\}: |F\circ \phi(y) -L | \lt \varepsilon.$
We know that there is a deleted neighbhorhood $[b-\alpha,b+\alpha]\setminus \{b\}$ where $\phi$ is continuous at all contained points.
Between these two deleted neighborhoods, we know that one must span a smaller distance (i.e. $\eta \leq \alpha$ or $\alpha \leq \eta$)...and note that if $\eta \leq \alpha$, then $\phi$ must also be continuous on $[b-\eta,b+\eta]\setminus\{b\}$; let us denote this 'smaller' deleted neighborhood as $[b-\mathcal \tau,b+ \mathcal \tau]\setminus\{b\}$.
Consequently, if we restrict $\phi$'s domain to $[b-\mathcal \tau,b+ \mathcal \tau]\setminus\{b\}$, $\phi$ must take on a minimum $m$ and a maximum $M$, and, by assumption, all values between $m$ and $M$, though possibly not $\phi(b)=a$. Let $\delta=\min(|a-m|,|a-M|)$. Then, for any $x \in (a-\delta, a+\delta) \setminus \{a\}$, there is necessarily a $y \in [b-\eta,b+\eta]\setminus\{b\}$ such that $\phi(y)=x$. But we know that $|F\circ \phi (y)-L| \lt \varepsilon$...or, equivalently, that $|F (x) - L| \lt \varepsilon$.