Let $E$ be a dense subset of a metric space $X$, and let $f$ be a uniformly continuous function from $E$ to complete metric space $Y$. For each $p\in X$, and each positive integer $n$, let $V_n(p)$ be the set of all $q\in E$ with $d(p,q)<\frac1n$. The diameter of closure of $f(V_n(p))$ is same as diameter of $f(V_n(p))$ itself, so these diameters tend to zero.
I don't understand why $\lim\limits_{n\to \infty} \text{diam} f(V_n(p))=0$. I thought this is because $V_{n+1}(p)\subset V_n(p)$, but that alone does not guarantee limit $0$.
Suppose otherwise. Then you'll have pairs of points which are very close to one another such the distances between their images is greater than a certain (fixed) distance. This goes against uniform continuity.