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.
 A: Your basic idea works; you just have to be a little more careful.
Since $X$ is a compact metric space, you can show that each $K_n$ has finite diameter. Moreover, since the sets $K_n$ are compact, there are points $x_n,y_n\in K_n$ such that $d(x_n,y_n)=\operatorname{diam}K_n$. The sequence $\langle x_n:n\in\Bbb Z^+\rangle$ has a convergent subsequence $\langle x_{n_k}:k\in\Bbb Z^+\rangle$, and the sequence $\langle y_{n_k}:k\in\Bbb Z^+\rangle$ has a convergent subsequence $\langle y_{n_{k_i}}:i\in\Bbb Z^+\rangle$. Let $x_0=\lim_ix_{n_{k_i}}$ and $y_0=\lim_iy_{n_{k_i}}$.
Let $i\in\Bbb Z^+$; then $x_{n_{k_j}}\in K_{n_{k_i}}$ for each $j\ge i$, and $K_{n_{k_i}}$ is closed, so $x_0\in K_{n_{k_i}}$. Thus, $x_0\in\bigcap_{i\in\Bbb Z^+}K_{n_{k_i}}=\bigcap_{n\in\Bbb Z^+}K_n=\{x\}$, and hence $x_0=x$. Essentially the same argument shows that $y_0=x$. Thus, $\lim_ix_{n_{k_i}}=x=\lim_iy_{n_{k_i}}$.
Let $\epsilon>0$; there is an $i\in\Bbb Z^+$ such that $d(x_{n_{k_j}},x)<\epsilon$ and $d(y_{n_{k_j}},x)<\epsilon$ for all $j\ge i$, and therefore
$$\operatorname{diam}K_{n_{k_j}}=d(x_{n_{k_j}},y_{n_{k_j}})\le d(x_{n_{k_j}},x)+d(x,y_{n_{k_j}})<2\epsilon$$
for each $j\ge i$. This shows that $\langle\operatorname{diam}K_{n_{k_i}}:i\in\Bbb Z^+\rangle\to 0$ as $i\to\infty$.
To complete the proof, just notice that since $K_{n+1}\subseteq K_n$ for each $n\in\Bbb Z^+$, $\operatorname{diam}K_{n+1}\le\operatorname{diam}K_n$ for each $n\in\Bbb Z^+$; the sequence $\langle\operatorname{diam}K_n:n\in\Bbb Z^+\rangle$ is therefore monotonically non-increasing, so the fact that it has a subsequence converging to $0$ implies that it also converges to $0$.
A: Let $\epsilon > 0$.  Then $Q = X - B_\epsilon(x)$ is a compact set.  The sets $X - K_n$ cover $Q$ so they ahve a finite subcover; in this case we have for some
$n\in\mathbb{N}$ $X - B_\epsilon(x) \subseteq X - K_n$.  Hence $K_n \subseteq B_\epsilon(x)$.
A: 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
$$\lim_{n \to \infty} \mathrm{diam} \ K_n =0$$
