Prove that $(K,\delta)$ is a compact metric space. Let $(X, d)$ be a compact metric space. For $x ∈ X $ and $\epsilon > 0$, define
{$B_{\epsilon}(x) := {y ∈
X | d(x, y) < \epsilon}$}.
For $C ⊆ X$ and $\epsilon > 0$, define $B_{\epsilon}(C) := ∪_{x∈C}B_{\epsilon}(x)$.
Let K be the
set of non-empty compact subsets of X. For $C, C_0 ∈ K$, define $δ(C, C_0) = $inf{$\epsilon| C ⊆ B_{\epsilon}(C_0)$ and $C_0 ⊆ B_{\epsilon}(C)$}.
Assuming that it forms a metric space show that $(K, δ) $ is a compact metric space.
My attempt:
The metric space $(X,d)$ is compact so it is totally bounded.
Let $B=${$C_k$ } be a sequence of compact metric spaces in $K$. We need to show that the infinite set has a limit point.
$(X, d)$ is totally bounded so let $A_1$={$x_{1,1},x_{2, 1},....,x_{k_1,1}$} such that ${\cup {B_d(x_i,1) }} $ covers X for $i= (1,1),..,(k_1,1)$ and ${\cup {B_d(x_i,\frac{1}{2}) }} $ covers X for $i= (1,2),..,(k_2,2)$ ,so $A_2= x_{1,2},x_{2,2},...,x_{k_2,2}$
So we can generate a new sequence of compact metric space $A_1,A_2,...,A_n$ as mentioned above.Now,let {$\cup {(A_i)}$}=$A$ .We take the closure $cl(A)$ which is the closed subset of $(X,d)$.Hence it is compact so it is in $(K,\delta)$.

$(B_{\delta}(cl(A) \cap B)  /cl(A))\ne \phi$ for all $\epsilon>0$ is what we have to show to prove that $cl(A)$ is the limit point of the sequence {${C_k}$} to show that the sequence has bolzano weistrass property.

We pick an $\epsilon >0$ and proceed as ,
Now, $C_i \subset B_{\frac{1}{N_1}}(cl(A))$ where $\frac{1}{N_1} < \frac{\epsilon}{2}$ and $C_i \subset B_{\epsilon}(cl(A))$ where $C_i$ is any compact subset in the sequence {$C_k$} and $C_i$ can be chosen to be different from $cl(A)$.
Now, let $x \in Cl(A)$
then $x \in X$ and $x \in \cup B_d(x_i,\frac{1}{N_1})$ where $i=(1,N_1),(2,N_1),(3,N_1),...,(k_{N_1},N_1)$
then $d(x,x_i) < \frac{1}{N_1}$ and let $x_n \in C_i$ then $d(x_i,x_n) < \frac{1}{N_1}$ so $d(x,x_n) < \epsilon$ and the other case should also follow when $ x = x_i $ where $i=(1,N_1),(2,N_1),(3,N_1),...,(k_{N_1},N_1)$.
So $cl(A) \subset B_{\epsilon}(C_i)$.
Then $\delta(C_i,cl(A)) < \epsilon$ which allows me to conclude that $cl(A)$ is the limit point of the above mentioned seqeunce (as $\epsilon$ can be chosen arbitrarily).So by bolzano property I can claim the compactness.
Is my attempt ok? I have tried hard so that the notations are understandable. Thanks in advance.
 A: I will write here my idea in a more formal way, and see if it works. I will try to do as little change in your argument as possible.
Idea: Lets start with something easier. Prove that for any convergent sequence $\{ C_i\}_{i=0}^\infty$ the limit is in $K$. THat is, $K$ is closed.
First, we would like to know how the limit of a sequence of compact spaces would look like. We look at some special cases for intuition.
In the case that $C_{i+1}\subseteq C_i$ it makes sense that the limit is $\bigcap C_i$.
Next, we look at the interval [0,1] and the sequence $[0,\frac{1}{2}], [\frac{1}{2},\frac{1}{2}+\frac{1}{4}], [\frac{1}{2}+\frac{1}{4},\frac{1}{2}+\frac{1}{4}+\frac{1}{8}]... $ and so on. What is the limit here? it seems that it should be $\{1\}$ (can you prove it?) But no matter what it is, the points of the $C_i$ has to get "closer and closer" to the points of the limit set.
Take $A_i$ as a set of points (finite) such that $\bigcup\limits_{j=1}^{k_i} B_d(x_{j,i},\frac{1}{i})$ covers $C_i$. With this sets, we can construct a sequence by first taking the elements of $A_1$ (ordered by the first index), then the elements of $A_2$ and so on. This is a sequence of elements of $X$, and as $X$ is a compact metric space, it will have (at least one) converging subsequences. So it makes sense to look at the set $C$ of the limit points of those subsequences.
Why this set is a good candidate for the limit? First if you look at the examples above, it coincides with the limit, or at least our idea of what the limit should be. Second, the set $C$ does not change if you change a finite set of the $C_i$'s, and most important (if  I am not wrong) it does not change if you choose any other set of points as your $A_i$'s (if the $C_i$'s are fixed).
Now, we have to prove that $C$ is compact. It is enough to prove that is closed. Try to show it, I will complete it later if you are not able to do it.
If all of this works. What we have left is to prove that $K$ is totally bounded. I don't see an easy proof of that right now but maybe you can do it.
A: $X$ is compact, so it is totally bounded; for each $n\in\Bbb N$ let $D_n$ be a finite $2^{-n}$-net in $X$, i.e., a finite subset of $X$ such that $X=\bigcup_{x\in D_n}B_d(x,2^{-n})$, and let $\mathscr{D}_n$ be the family of non-empty subsets of $D_n$.
Let $\epsilon>0$; there is an $n\in\Bbb N$ such that $2^{-n}<\epsilon$. For $C\in K$ let
$$F=D_n\cap B_d(C,2^{-n})\in\mathscr{D}_n\,;$$
then $\delta(C,F)\le 2^{-n}<\epsilon$, so $\mathscr{D}_n$ is a finite $\epsilon$-net for $K$, and $K$ is totally bounded. To show that $K$ is compact, it only remains to show that $K$ is complete.
Let $\langle C_n:n\in\Bbb N\rangle$ be a Cauchy sequence in $K$. Let
$$H=\bigcap_{n\in\Bbb N}\operatorname{cl}_X\left(\bigcup_{k\ge n}C_k\right)\in K\,,$$
and let $\epsilon>0$. If $\{n\in\Bbb N:C_n\nsubseteq B_d(H,\epsilon)\}$ is infinite, then $\bigcup_{k\ge n}C_k\nsubseteq B_d(H,\epsilon)$ for each $n\in\Bbb N$, so $\operatorname{cl}_X\left(\bigcup_{k\ge n}C_k\right)\setminus B_d(H,\epsilon)$ is a non-empty compact set for each $n\in\Bbb N$, and
$$H\setminus B_d(H,\epsilon)=\bigcap_{n\in\Bbb N}\left(\operatorname{cl}_X\left(\bigcup_{k\ge n}C_k\right)\setminus B_d(H,\epsilon)\right)\ne\varnothing\,,$$
which is absurd. Thus, for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $C_n\subseteq B_d(H,\epsilon)$ for all $n\ge n_\epsilon$.
Now suppose that $\{n\in\Bbb N:H\nsubseteq B_d(C_n,\epsilon)\}$ is infinite. There is an $m\in\Bbb N$ such that $\delta(C_k,C_\ell)<\frac{\epsilon}3$ whenever $k,\ell\ge m$. Fix $n\in\Bbb N$ such that $n\ge m$ and $H\nsubseteq B_d(C_n,\epsilon)$, and let $x\in H\setminus B_d(C_n,\epsilon)$. Let $k\ge n$. Then $\delta(C_k,C_n)<\frac{\epsilon}3$, so $C_k\subseteq B_d\left(C_n,\frac{\epsilon}3\right)$, and hence $B_d\left(C_k,\frac{\epsilon}3\right)\subseteq B_d\left(C_n,\frac{2\epsilon}3\right)$. But then
$$\begin{align*}
H&\subseteq\operatorname{cl}_X\left(\bigcup_{k\ge n}C_k\right)\subseteq\operatorname{cl}_X\left(\bigcup_{k\ge n}B_d\left(C_k,\frac{\epsilon}3\right)\right)\\
&\subseteq\operatorname{cl}_X\left(B_d\left(C_n,\frac{2\epsilon}3\right)\right)\subseteq B_d(C_n,\epsilon)\\
&\subseteq H\setminus\{x\}\,,
\end{align*}$$
which is absurd.
Thus, there is an $m_\epsilon$ such that $H\subseteq B_d(C_n,\epsilon)$ for all $n\ge m_\epsilon$.
Putting the pieces together, we see that $\delta(H,C_k)\le\epsilon$ for all $k\ge\max\{n_\epsilon,m_\epsilon\}$ and hence that $\langle C_n:n\in\Bbb N\rangle$ converges to $H$ in $\langle K,\delta\rangle$.
