This is a theorem that was discussed in class but I came up with another proof.
The theorem is: Let $X$ and $Y$ be metric spaces. If $X$ is compact, and $h$ is a set of mappings from $X$ to $Y$ and is equicontinuous everywhere in $X$, then $\forall\epsilon>0,\exists\delta>0,s.t.\forall f\in h,x,x'\in X(d(x,x')<\delta\Rightarrow d(f(x),f(x')<\epsilon)$.
My proof is as the following:
Fix $\epsilon$.
Let $\phi(x_0):=\max{\lbrace \delta>0|\forall f\in h,x\in X(d(x_0,x)<\delta\Rightarrow d(f(x_0),f(x)<\epsilon)\rbrace}$, then since $\lim_{x\rightarrow x_0}\phi(x)=\phi(x_0)$, we know that $\phi$ is a continuous mapping.
Since compact sets in metric space are closed sets, $X$ is a closed set, which means $\phi(X)$ is a closed set on $\mathbb{R}$, so $\min \phi(X)$ exists, and any $\delta$ satisfies $\delta < \min \phi(X)$ is the $\delta$ we find.
Is this correct?
Your proposed proof reminds me of this proof of the Lebesgue number lemma (in fact, a common proof of your desired proposition is proven using the Lebesgue number lemma -- see Equicontinuous on a compact set implies uniform equicontinuous).
I think that essentially the same technique used to prove $r(x)$ is continuous in the first answer I linked, can show that your $\phi(x)$ is continuous. Indeed, fixing $\epsilon>0$, we can define $$U := \bigcup_{x_0 \in X} \{x \in X: d_Y(f(x),f(x_0))<\epsilon, \; \forall f\in \cal F\}$$ which is an open set because it is the union of open sets (the sets on the RHS are open I think by equicontinuity?), and your $\phi(x_0)$ is exactly $r(x_0,U):= \sup\{\delta>0: B(x_0,\delta) \subseteq U\}$.
For a brief proof sketch that $r$ is continuous, Ctrl-F "Lebesgue number lemma" in this pdf, and read the paragraph preceding the light-blue lemma box.