Let $X$ be a complete metric space and $\varnothing \neq A_{n}\subseteq X$ closed for all $n \in \mathbb N$ so that $A_{n+1} \subseteq A_{n}$ and $\operatorname{diam{(A_{n})}}\xrightarrow{n \to \infty}0$. Show there exists unique $x \in X$ so that $\bigcap\limits_{n \in \mathbb N} A_{n} = \{x\}$.
My idea: We can construct a cauchy sequence $(x_{n})_{n}$ where $x_{n} \in A_{n}$ and $x_{m} \in A_{m}$ so that, for any $\epsilon > 0$ there is a $N \in\mathbb N$ so that for any $n>m \geq N$: we get $d(x_{m}, x_{n})\leq \operatorname{diam{(A_{N})}}<\epsilon$ thus $(x_{n})_{n}$ is a cauchy sequence and for completeness of $X$, there exists limit point $x \in X$. In order to show that $x$ is indeed unique. We assume $\exists y\neq x \in X$ that also lies in $\bigcap\limits_{n \in \mathbb N} A_{n}$, since we are in a metric space, $X$ is Hausdorff and thus there exists balls $B_{\delta_{y}}(y), B_{\delta_{x}}(x)$ so that $B_{\delta_{y}}(y) \cap B_{\delta_{x}}(x)=\varnothing$ but this immediately leads to a contradiction as we choose $N\in \mathbb N$ so that $\operatorname{diam{(A_{N})}}<\min{\{ \delta_{y}, \delta_{x}\}}$ but we already know that $x\in A_{N}$ and thus $y\notin A_{N}\Rightarrow y \notin \bigcap\limits_{n \in \mathbb N}A_{n}$. Is my proof, particularly the second part ok?
Example style. Not meant as a criticism of the OP.
(1). Take some $a_n\in A_n$ for each $n.$
(2). For $\epsilon >0$ take $m$ such that diam($A_m)<\epsilon$. Then for all $n,n'$ we have $$n,n'\ge m \implies \{a_n,a_{n'} \}\subset A_m \implies d(a_n,a_{n'})\le \operatorname{diam}(A_m)<\epsilon.$$
So the sequence $ (a_n)_n$ is Cauchy. And $d$ is complete so $(a_n)_n$ converges to some $x\in X$.
(3). For each $n$ we have $x\in \overline {\{a_m:m\ge n\}}\subset \overline {A_n}=A_n .$ So $x\in \cap_nA_n.$
(4). For all $y$ we have $$y\in \cap_n A_n \implies \forall n\,(\,\{x,y\}\subset A_n\,) \implies$$ $$\implies \forall n\,(\,d(x,y)\le \operatorname{diam}(A_n)\,)\implies$$ $$\implies d(x,y)\le \inf_n \operatorname{diam}(A_n)=0\implies$$ $$\implies d(x,y)=0\implies y=x.$$
(5). By (3) and (4) we have $\{x\}\subset \cap_n A_n\subset \{x\}.$ Therefore $\cap_n A_n=\{x\}.$