Let $(X,d)$ a complete metric space. I need to prove that $(\mathcal{CB}(X),d_H)$ is complete. I see other posts (yes, it's a duplicate but no one answered this question) and the only thing I couldn't do is to prove that if $\{A_n\}_{n\in\mathbb{N}}\subseteq\mathcal{CB}(X)$ is a Cauchy's decreasing sequence on $(\mathcal{CB}(X),d_H)$ and $A=\bigcap_{n=1}^{\infty}{A_n}$ (I already know that $A\neq \varnothing$ and, it's closed and bounded), then $\lim_{n\to\infty}{d_H(A_n,A)}=0$.
Fixed $\varepsilon>0$, since $A\subseteq A_n(\varepsilon)$ for all $n\in\mathbb{N}$, I just need to find $n_0=n_0(\varepsilon)$ such that $A_{n_0}\subseteq A(\varepsilon)$. In the space $\mathcal{K}(X)$ of non-empty compact sets of $X$ is easier to prove by contradiction (there's a post about it), but I don't know how to adapt the proof to the space of closed and bounded sets of $X$.
Partial answer.
As far as I can tell, all you have left to do is prove that $$\lim_{n\to\infty} d_{\mathrm H}(A_n,A)=0$$ We know that, since $(A_k)_{k\in\mathbb N}$ is a decreasing sequence, that $$A_{k+1}\subset A_k$$
With this in mind let $x_n=d_{\mathrm H}(A_n,A)$. We will show $(x_n)_{n\in\mathbb N}$ is decreasing.
We define the Hausdorff metric as $$d_{\mathrm H}(A,B)=\max\big(\operatorname{dist}(A,B),\operatorname{dist}(B,A)\big) \\ \operatorname{dist}(A,B)=\sup_{a\in A}\inf_{b\in B}d(a,b)$$
So $$x_{n+1}=\max\big(\operatorname{dist}(A_{n+1},A)~,~\operatorname{dist}(A,A_{n+1})\big)$$
Well, you can see that $$\operatorname{dist}(A_{n+1},A)=\sup_{p\in A_{n+1}}\inf_{q\in A}d(p,q)$$ However we know that $A_{n+1}\subset A_{n}$, hence $$\sup_{p\in A_{n+1}}(\cdots )\leq \sup_{p\in A_n}(\cdots)$$ This is because, if $X\subseteq Y$, then $\sup X\leq \sup Y$. Therefore, $$\operatorname{dist}(A_{n+1},A)\leq \operatorname{dist}(A_{n},A)$$
What about the other argument?
$$\operatorname{dist}(A,A_{n+1})=\sup_{p\in A}\inf_{q\in A_{n+1}}d(p,q)$$
But this is of course zero. Note that, using the same simple fact as before, $$0\leq \sup_{p\in A}\inf_{q\in A_{n+1}}d(p,q)\leq \sup_{p\in{A_{n+1}}}\inf_{q\in A _{n+1}}d(p,q) \\ 0\leq \sup_{p\in A}\inf_{q\in A_{n+1}}d(p,q)\leq \operatorname{dist}(A_{n+1},A_{n+1}) \\ 0\leq \sup_{p\in A}\inf_{q\in A_{n+1}}d(p,q)\leq 0 \\ \implies \sup_{p\in A}\inf_{q\in A_{n+1}}d(p,q)=0$$
So, $$x_{n}=\max\big(\operatorname{dist}(A_{n},A)~,~0\big) \\ x_{n+1}=\max\big(\operatorname{dist}(A_{n+1},A)~,~0\big)$$
But, if $a<b$, then $\max(a,0)\leq \max(b,0)$, so since $\operatorname{dist}(A_{n+1},A)\leq \operatorname{dist}(A_{n},A)$ we get $x_{n+1}\leq x_n$.
So, since $(x_n)_{n\in\mathbb N}$ is decreasing and bounded below by $0$, it is convergent. Now you just need to show that it converges exactly to $0$.