Proving that a certain space of compact sets is a complete metric space 
Let $(X,d)$ be a metric space, and let $K(X)$ denote the collection of all non-empty compact subsets of $X$. Define a function, $d_h\colon K(x)\times K(x)\to\mathbb R$ by letting $$d_h(A,B)=\inf\{\varepsilon\colon A\subseteq U_\varepsilon(B) \text{ and } B\subseteq U_\varepsilon(A)\},$$ where $ U_\varepsilon(S)=\{x\in X\colon d(x, S)<\varepsilon\} $, and the distance $d(x, S)$ from a point $x\in X$ to a non-empty subset $S$ of $X$ is defined by $d(x,S)=\inf\{d(x,s)\colon s\in S\}$.
Prove that $(K(X), d_h)$ is a complete metric space.


My attempt:
First, we need to show that $d_h$ is a metric. It is trivial to show that $d_h\ge 0$ and $d_h=0$ if and only if $d_h=0$. Also, it is clear that $d_h(A,B)=d_h(B,A)$.
To see that for every $A, B, C\in K(X)$, we have $d_h(A,C)\le d_h(A,B)+d_h(B,C)$, we consider $\gamma_{A}:=\{x\colon d(x,A)=d_h(A,B)\}$ and $\gamma_{B}:=\{x\colon d(x,B)=d_h(B,C)\}$.
If $d_h(A,C)\le d_h(A,B)$ or $d_h(A,C)\le d_h(B,C)$ then we are through, otherwise we claim that $$d_h(B,C)\ge d_h(A,C)-d_h(A,B).$$ Suppose the converse, $\gamma_B$ would be completely lying in the interior of $\gamma_A$. By the definition of $d_h(B,C)$, $C$ would be lying in $ U_{d_{h}(B,C)}$ and in the interior of $$\gamma:=\{x\colon d(x,A)=d_h(B,C)+d_h(A,B)<d_h(A,C)\}.$$ Interchanging the roles played by $A$ and $C$, we will have a contradiction with the definition of $d_h(A,C)$. Hence $d_h$ is a metric.
However, I got stuck in proving completeness. Suppose we have a sequence $\{C_n\}_{n=1}^\infty$ with $C_n\in  K(X)$ and $d_h(C_n,C_m)\to 0$ as $n,m\to\infty$, I failed to find out the limit set. I think it should be $$\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty C_k$$ but I am not even sure whether it is compact or not. Any help here? Thank you!

Edit: As user10354138 pointed out, we should assume $X$ is a complete metric space.
 A: You need to assume $X$ is complete, otherwise any Cauchy sequence $x_n\in X$ not converging to a limit gives you a corresponding sequence $\{x_n\}\in K(X)$ that does not converge to a limit.
After that, you need to stick a closure (again, see the Cauchy sequence example)
$$
C_\infty=\bigcap_{n=1}^\infty \overline{\bigcup_{k=n}^\infty C_k}
$$
The first nonobvious point is $\overline{\bigcup_{k\geq n}C_k}$ is compact: it is a closed subset of the complete space $X$, so it suffices to show total boundedness.  So we need to show we can cover with finitely many $2\varepsilon$-balls, which using $(C_k)$ is Cauchy we reduce to covering the finitely many initial $C_k$ by $2\varepsilon$-balls, plus covering $C_m$ with $\varepsilon$-balls where all $C_n$s, $n>m$, lie within $\varepsilon/2$ of $C_m$.  Now enlarge the $\varepsilon$-balls to $2\varepsilon$-balls and we covered the lot.
So $C_\infty$ is nested intersection of compacts so is nonempty compact.  We also know
$$
d_h(C_n,\overline{\bigcup_{k\geq m} C_k})
\leq\sup_{k\geq m} d_h(C_n, C_k)\xrightarrow{\min(m,n)\to\infty} 0
$$
So we may assume, by replacing $C_n$ with $\overline{\bigcup_{k\geq n} C_k}$, that $(C_n)$ are nested.  In this case the proof of $d_h(C_n,C_\infty)\to 0$ just need to bound $\sup_{x\in C_n} d(x,C_\infty)$ (the $\sup_{x\in C_\infty} d(x,C_n)=0$ come from nested): for $\varepsilon>0$, pick sequence $n_k$ such that $d_h(C_m,C_n)<\varepsilon/2^k$ for all $m,n\geq n_k$.  Then every point $x_1\in C_{n_1}$ is less than distance $\varepsilon/2$ from a point $x_2\in C_{n_2}$, which in turn is less than distance $\varepsilon/2^2$ from $x_3\in C_{n_3}$, etc. and $x_n$ is Cauchy, so $x_n\to x\in C_\infty$, with $$d(x,x_1)<\varepsilon/2+\varepsilon/2^2+\dots=\varepsilon.$$
