Given $A \subseteq \mathbb R^n$, if $G \colon A \to \mathbb R^m$ is a uniformly continuous function, then given $\epsilon$ there is $\delta$ so that if $B\subseteq A$ and $\operatorname{diam}(B)<\delta$, then $\operatorname{diam}(F(B))< \epsilon$.
Isn't the definition of uniformly continuous that $||F(x)-F(y)|| < \epsilon$ whenever $x,y\in D$ and $||x-y||<\delta$? How do I relate this with $\operatorname{diam}(A)$ and the fact that $G$ is uniformly continuous? I am very new to these ideas.
For all $\varepsilon >0$, pick $\delta >0$ such that $$||x-y|| \leq \delta \Rightarrow ||F(x) - F(y)|| \leq \frac{\varepsilon}{3}$$
This can be done since $F$ is uniformly continuous.
Now, suppose $B$ is any subset of the domain of $F$ whose diameter is $diam(B) \leq \delta$. Then $$d:=diam(F(B)) = \sup_{x,y \in B}||F(x)-F(y)||$$ using the definition of $\sup$ you can find $x,y \in B$ such that $||F(x)-F(y)|| + \frac{\varepsilon}{3} \geq d$. But $diam(B) \leq \delta$, so $||x-y|| \leq \delta$. So finally $$d \leq ||F(x)-F(y)|| + \frac{\varepsilon}{3} \leq \frac{\varepsilon}{3} + \frac{\varepsilon}{3} < \varepsilon$$