Suppose we have a uniformly continuous function $f$ on some metric space $X$ and the nested sets $V_n(p)$, where $V_n(p)$ is the neighborhood of radius $1/n$ about $p$ and $n \in \mathbb{N}$. I am trying to understand the following claim: $\textrm{diam}\,f(V_n(p)) \rightarrow 0$ as $n\rightarrow \infty$.
Intuitively it makes sense that as the diameter of $V_n(p)$ shrinks, so should the diameter of its image. However, I do not see how the definition of uniform convergence implies that it must approach zero.
Fix $\varepsilon>0$. There is some $\delta>0$ such that $d(f(x),f(y))<\varepsilon$ whenever $d(x,y)<\delta$. Now consider $n_0$ such that $n^{-1}_0<\delta$. Then $\operatorname{diam}f(V_n(p))\leqslant \varepsilon$ for all $n\geqslant n_0$.
Note that actually we just need $\delta$ such that $d(f(x),f(p))<\varepsilon$ whenever $d(x,p)<\delta$ (we will get $\operatorname{diam}f(V_n(p))\leqslant 2\varepsilon$, so continuity is enough.