Let $X$ be a metric space, $\mathcal{F}$ and $\mathcal{G}$ are families of real valued functions. Suppose that $\mathcal{F}$ and $\mathcal{G}$ are equicontinuous. I ever read that union of two equicontinuous families of functions is equicontinuous, but the book was not given the proof.
Then, I want to prove it. First, I think union of two equicontinuous families of functions can be written like this, $\mathcal{F}\cup \mathcal{G}=\{h: h\in \mathcal{F}\, \text{or}\, h\in\mathcal{G} \}$
To prove it is equicontinuous, let $\varepsilon>0$. Since $\mathcal{F}$ is equicontinuous, then there exists $\delta_1>0$, such that for all $x,y\in X$ with $d(x,y)<\delta_1$,we have $|f(x)−f(y)|<\varepsilon$, for all $f\in\mathcal{F}$.
And since $\mathcal{G}$ is equicontinuous, then there exists $\delta_2>0$, such that for all $x,y\in X$ with $d(x,y)<\delta_2$,we have $|g(x)−g(y)|<\varepsilon$, for all $g\in\mathcal{G}$.
Choose $\delta=\max\{\delta_1,\delta_2\}$, such that for all $x,y\in X$ with $d(x,y)<\delta$, we have $|h(x)-h(y)|<\varepsilon$, for all $h\in \mathcal{F}\cup\mathcal{G}$.
Is there any mistake on the proof? I am not sure about the $\delta$ that I choose. Thanks for any advice.
Correction:By comment from @geetha290krm, the correct $\delta$ that should I choose is $\delta=\min\{\delta_1,\delta_2\}$, such that for all $x,y\in X$ with $d(x,y)<\delta$, we have $|h(x)-h(y)|<\varepsilon$, for all $h\in \mathcal{F}\cup\mathcal{G}$.