I am trying to prove:
Show that the requirement in the definition of uniform continuity can be rephrased as follows: To every $\epsilon > 0, \exists \delta > 0$ such that $\forall E \subset X \textrm{ with } \operatorname{diam} E < \delta, \textrm{ we have that } \operatorname{diam} f(E) < \epsilon$.
My attempt:
($\Rightarrow$) Let $f: X \to Y$ be a uniformly continuous function and $\epsilon > 0$. By the definition of uniform continuity, we know that $\exists \delta > 0$ such that for any $x, y \in X$, we have \begin{equation}\tag{9.1} d_X(x, y)<\delta \implies d_Y(f(x), f(y))<\frac{\epsilon}{2} \end{equation} Now, suppose $E \subset X$ is such that $\text{diam}(E)<\delta$. Then, for all $x, y \in E$, we have $d_X(x, y) \le \text{diam}(E)<\delta$, so that from (9.1) we have $d_Y(f(x), f(y))<\dfrac{\epsilon}{2}$. Note that $\dfrac{\epsilon}{2}$ effectively acts as an upper bound for the set $S = \{d(f(x), f(y))\mid x,y\in E\}$. Since $\text{diam}(f(E))$ is the least upper bound (or supremum) of $S$ by definition, we have that $$\text{diam}(f(E))=\sup_{x,y \in E}d(f(x),f(y))\leq \epsilon/2<\epsilon$$ and we are done.
($\Leftarrow$) Let $\epsilon>0$ be given. Suppose that $\exists \delta > 0$ such that $\forall E \subset X \textrm{ with } \operatorname{diam} E < \delta$, we have that $\operatorname{diam} f(E) < \epsilon$. Let $E \subset X$ be such that $\operatorname{diam} E < \delta$. Let $x, y \in E$. Then it must be that $d_X(x, y) \le \operatorname{diam} E < \delta$. Note that since $x$ and $y$ were arbitrarily chosen members of $E$, our $\delta$ is independent of $x$ (and $y$). From our hypothesis, it follows that $d_Y(f(x), f(y)) \le \operatorname{diam} f(E) <\epsilon$. Therefore, $f: X \to Y$ is a uniformly continuous function.
Can someone please verify my proof and let me know if it can be improved in some manner? I am afraid that the second half of the proof is not very clear; for instance, I don't think the standard implication-style structure of the definition of uniform continuity is clear from how I have worded the second half of my proof.
Your proof of $\implies$ is correct but the converse part is not correct. You just took some $E$ with diameter less than $\delta$ and showed that $d(f(x),f(y)) <\epsilon$ for $x ,y \in E$. This does not prove uniform continuity. For a correct proof simply take $\delta$ as in the hypothesis and take $E=\{x,y\}$. (This choice of $E$ is important for the proof). If $d(x,y) <\delta /2$ then diameter of $E$ is less than $\delta$ and this gives $d(f(x),f(y)) <\epsilon$.