This is a question about the proof of Theorem 3.10(b) in baby Rudin:
For any $E \subset X$ note $\mathrm{diam} \; E = \sup \big\{ d(p, q) : p, q \in E \big\}$ and let $\{K_n\}$ be a collection of nonempty compact sets in a metric space $X$ such that for any $n \in \mathbb{N}$, $K_n \supset K_{n+1}$.
If $\lim_{n \to \infty} \mathrm{diam} \; K_n = 0$, then $\bigcap_{n=1}^\infty K_n$ contains exactly one element.
If $\bigcap_{n=1}^\infty K_n$ were to contain more than one element, then we would have: $\mathrm{diam} \; K_n \geq \mathrm{diam} \;\bigcap_{n=1}^\infty K_n > 0$ for every $n \in \mathbb{N}$. In Rudin's proof, he says that this contradicts the fact that $\mathrm{diam} \; K_n \to 0$.
But how is this a contradiction? For example, if $\mathrm{diam} \; K_n = \frac{1}{n}$ for every $n \in \mathbb{N}$, we would have a sequence which converges to zero, but each of its elements is $> 0$.