Proving that if $\bigcap_{n=1}^{\infty }K_{n}=\left \{ x \right \}\Rightarrow\operatorname{diam}(K_{n})\rightarrow 0$ let $(X,d)$ be a compact metric space if $K_{n}={}$closed and $K_{n+1}\subseteq K_{n} $
and $\bigcap_{n=1}^{\infty }K_{n}=\left \{ x \right \}$ show that $\operatorname{diam}(K_{n})\rightarrow 0$
I know there exists a different proof,  but my first thought was to prove this with subsequences which is probably wrong, I wonder whether  there is any way we can prove this with subsequences?
my attempt:
I think we can approach $\operatorname{diam}(K_{n})$ with subsequences $d(x_{k_{n}},y_{k_{n}})\rightarrow \operatorname{diam}(K_{n})$
know  $x,y\epsilon K_{n}$ such that $x_{k_{n}} \rightarrow x$ and $y_{k_{n}} \rightarrow y$
$d(x_{k_{n}},y_{k_{n}})\leq d(x_{k_{n}},x)+d(x,y)+d(y,y_{k_{n}})$ and since $\bigcap_{n=1}^{\infty }K_{n}=\left \{ x \right \}$
and $n \to \infty $ $x=y$ $\Rightarrow $
$\operatorname{diam}(K_{n})\leq 0$
I think I know where the mistake is. I just need to fully understand why this can't work ,
if there is a way to prove this sequences, I would appreciate if someone could write it below.

Here is a proof.
For all $\varepsilon >0$, call
$$H_n= K_n \setminus B(x, \varepsilon )$$
where $B(x, \varepsilon )$ is the open ball of radius $\varepsilon$ centered at $x$.
Clearly $H_n$ is compact and $H_{n+1} \subseteq H_n$.
Note that
$$\bigcap_n H_n = \bigcap_n K_n \setminus B(x, \varepsilon ) = \left\{ x \right\} \setminus B(x, \varepsilon ) = \emptyset $$
This means that one of the $H_n$ is empty (if all were not-empty, then their intersection would be not-empty). Say $H_N$ is the first one to be empty (and so are also $H_{N+1}, H_{N+2}, ...$)
In other words $K_N \subseteq B(x, \varepsilon )$. In particular $$\mathrm{diam} \ K_N \le \mathrm{diam} \ B(x, \varepsilon ) = 2 \varepsilon$$
Hence, for all $\varepsilon >0$ there exists $N >0$ such that for all $n \ge N$
$$\mathrm{diam} \ K_n \le 2 \varepsilon$$
which is by definition
$$\lim_{n \to \infty} \mathrm{diam} \ K_n =0$$
