Let $f: (X, d) \to (Y, d')$. Prove that the following are equivalent:
a) $f$ is uniformly continuous in $X$.
b)For every pair of sequences $(x_n), (y_n) \subseteq X$ such that $ d(x_n, y_n) \to 0$ then $d'(f(x_n), f(y_n))\to .0$
c) For all $\varepsilon >0$, there exists $\delta>0$ such that for all $E \subset X$ with $\mbox{diam}(E)<\delta$ then $\mbox{diam}(f(E))<\varepsilon$
I was able to prove that a) implies b) and that c) implies a), however when trying to proove that b)implies c) I am having difficulties. I've done the following;
EDIT: Complete proof with some help.
Now if b) stands lets suppose that c) does not. Then $\exists \epsilon>0$ such that for all $\delta>0$ there exists $E\subset X$ such that $\mbox{diam}(E)<\delta$ and $\mbox{diam}(f(E))\geq \epsilon$.
Take $\delta = \frac{1}{n}$ then for each $n \in \mathbb{N}$ there exists $E_n \subset X$ such that $\mbox{diam}(E_n)<\frac{1}{n}$ and $\mbox{diam}(f(E_n))\geq \epsilon$.
That is, for each $n\in \mathbb{N}$, there exist $x_n,y_n \in E_n$ such that $d'(f(x_n),f(y_n)) \geq \frac{\epsilon}{2}$.
Making $n\to \infty$ we have that $d(x_n,y_n)\to 0$ and $d'(f(x_n),f(y_n)) \geq \frac{\epsilon}{2}$ which contradicts the hypothesis b).
You are almost done:
Assume (c) does not hold. Then there is $\epsilon >0$ such that:
For each $n$, there is $E_n$ so that diam $(E_n) < \frac 1n$ and diam$(F(E_n)) \ge \epsilon$.
Then there is $x_n, y_n \in E_n$ so that $d'(f(x_n), f(y_n)) \ge \epsilon/2$.