The subspace $C$ of the space $(X,d)$ is compact Let:

*

*$(X,d)$ - complete metric space.

*$K_1,K_2,\dots $ - compact sets in the space $(X,d)$ (sequence $K_1,K_2,\dots$ need not be descending)

*$C_n=\{x\in X: \text{dist} (x,K_n)\le \frac 1n \}$ for  $n=1,2,\dots$

*$C=\bigcap\limits_{n=1}^{\infty} C_{n}\subset X$
Claim: The subspace $C$ of the space  $(X,d)$ is compact
My attempts:
We want to show that $\bigcap\limits_{n=1}^{\infty} C_{n}$ is compact. Let's establish $V=\{ \bigcup\limits_{\alpha}:\alpha \in A\}$ - cover of $\bigcap\limits_{n=1}^{\infty} C_{n}$. From this $V$ we want to choose finite subcover for $\bigcap\limits_{n=1}^{\infty} C_{n}$. I think that we can use the total limitation of space $C$ with metric $d$ truncated to $C$. However I don't have idea how I can do it.
 A: We assume the following fact without proof:
If the topology of a topological space is induced by a metic, then
a set $K$ is compact iff it is sequentially compact (i.e., every
sequence in $K$ has a subsequence that converges to a point in $K$).

Claim 1: Let $C=\cap_{n}C_{n}$. For any sequence $(x_{n})$ in $C$
and any $\varepsilon>0$, there exists a subsequence $(x_{n_{k}})$
of $(x_{n})$ such that $d(x_{n_{k_{1}}},x_{n_{k_{2}}})<\varepsilon$
for all $k_{1},k_{2}\in\mathbb{N}$.
Proof of Claim 1: Let $(x_{n})$ be a sequence in $C$ and $\varepsilon>0$.
Choose $N_{1}\in\mathbb{N}$ such that $\frac{1}{N_{1}}<\frac{\varepsilon}{4}$.
Since $x_{n}\in C_{N_{1}}$, there exists $y_{n}\in K_{N_{1}}$ such
that $d(x_{n},y_{n})\leq\frac{1}{N_{1}}$. Since $K_{N_{1}}$
is compact, for the sequence ($y_{n})$, there exists a subsequence
$(y_{n_{k}})$ and $y\in K_{N_{1}}$ such that $y_{n_{k}}\rightarrow y$.
Choose $N_{2}$ such that $d(y_{n_{k}},y)<\frac{\varepsilon}{4}$
whenever $k\geq N_{2}$. Let $N=\max(N_{1},N_{2})$. Consider the
subsequence $(x_{n_{k}})_{k\geq N}$. Let $k_{1},k_{2}\geq N.$ We
have that
\begin{eqnarray*}
 &  & d(x_{n_{k_{1}}},x_{n_{k_{2}}})\\
 & \leq & d(x_{n_{k_{1}}},y_{n_{k_{1}}})+d(y_{n_{k_{1}}},y)+d(y,y_{n_{k_{2}}})+d(y_{n_{k_{2}}},x_{n_{k_{2}}})\\
 & \leq & \frac{1}{N_{1}}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{1}{N_{1}}\\
 & < & \varepsilon.
\end{eqnarray*}

Claim 2: Every sequence in $C$ has a convergent subsequence.
Proof of Claim 2: We prove by Cantor's diagonal argument. Let $(x_{n})$
be a sequence in $C$. Choose a subsequence $(x_{1,k})_{k}$ of $(x_{n})$
such that $d(x_{1,k_{1}},x_{1,k_{2}})<\frac{1}{1}$ for any $k_{1},k_{2}\in\mathbb{N}$.
Suppose that a subsequence $(x_{n,k})_{k}$ has been chosen. Invoking
Claim 1 for the sequence $(x_{n,k})_{k}$ and positive number $\varepsilon=\frac{1}{n+1}$,
we choose a subsequence $(x_{n+1,k})_{k}$ of $(x_{n,k})_{k}$ such
that $d(x_{n+1,k_{1}},x_{n+1,k_{2}})<\frac{1}{n+1}$ for any $k_{1},k_{2}\in\mathbb{N}$.
Define $y_{n}=x_{n,n}$. We go to show that $(y_{n})$ is a convergent
subsequence of $(x_{n})$. To be precise, we show all working steps
in detail. By choosing a subsequence $(x_{1,k})_{k}$ of $(x_{n})$,
we mean choosing a strictly increasing function $\theta_{1}:\mathbb{N}\rightarrow\mathbb{N}$
such that $x_{1,k}=x_{\theta_{1}(k)}$. Since $(x_{n+1,k})_{k}$ is
a subsequence of $(x_{n,k})_{k}$, there exists a strictly increasing
function $\theta_{n+1}:\mathbb{N}\rightarrow\mathbb{N}$ such that
$x_{n+1,k}=x_{n,\theta_{n+1}(k)}.$ Now,
\begin{eqnarray*}
 &  & x_{n,n}\\
 & = & x_{n-1,\theta_{n}(n)}\\
 & = & x_{n-2,\theta_{n-1}(\theta_{n}(n))}\\
 & = & \ldots\\
 & = & x_{1,\theta_{2}\circ\theta_{3}\circ\cdots\circ\theta_{n}(n)}\\
 & = & x_{\theta_{1}\circ\theta_{2}\circ\cdots\circ\theta_{n}(n)}.
\end{eqnarray*}
Define $\psi:\mathbb{N}\rightarrow\mathbb{N}$ by $\psi(n)=\theta_{1}\circ\cdots\circ\theta_{n}(n)$.
Observe that $\theta_{n+1}(n+1)\geq n+1>n$, so $(\theta_{1}\circ\cdots\circ\theta_{n})\left(\theta_{n+1}(n+1)\right)>(\theta_{1}\circ\cdots\circ\theta_{n})(n)$,
i.e., $\psi(n+1)>\psi(n)$. Therefore, $\psi$ is a strictly increasing
function. Since $y_n = x_{\psi(n)}$. This shows that $(y_{n})$ is a subsequence of $(x_{n})$.
Let $\varepsilon>0$ be given. Choose $N$ such that $\frac{1}{N}<\varepsilon$.
Let $n\geq N$ and $k\in\mathbb{N}$. Observe that
\begin{eqnarray*}
y_{n+k} & = & x_{n+k,n+k}\\
 & = & x_{n+k-1,\theta_{n+k}(n+k)}\\
 & = & x_{n+k-2,\theta_{n+k-1}\circ\theta_{n+k}(n+k)}\\
 & = & \cdots\\
 & = & x_{n,\theta_{n+1}\circ\theta_{n+2}\circ\cdots\circ\theta_{n+k}(n+k)}.
\end{eqnarray*}
Therefore
\begin{eqnarray*}
d(y_{n+k},y_{n}) & = & d(x_{n,\theta_{n+1}\circ\theta_{n+2}\circ\cdots\circ\theta_{n+k}(n+k)},x_{n,n})\\
 & < & \frac{1}{n}\\
 & < & \varepsilon.
\end{eqnarray*}
This shows that $(y_{n})$ is a Cauchy sequence and hence is convergent
because $(X,d)$ is complete.

Claim 3: If $K$ is compact, $l>0$, then $C_{K,l}=\{x\in X\mid d(x,K)\leq l\}$
is closed.
Proof of Claim 3: Let $(x_{n})$ be a sequence in $C_{K,l}$ and suppose
that $x_{n}\rightarrow x$ for some $x\in X$. We go to show that
$x\in C_{K,l}$. For each $n$, choose $y_{n}\in K$ such that $d(x_{n},y_{n})<d(x_{n},K)+\frac{1}{n}\leq l+\frac{1}{n}.$
Since $K$ is compact, for the sequence ($y_{n})$, there exists a
subsequence $(y_{n_{k}})$ such that $y_{n_{k}}\rightarrow y$ for
some $y\in K$. We have that $d(x_{n_{k}},y_{n_{k}})<l+\frac{1}{n_{k}}$.
Letting $k\rightarrow\infty$, we have $d(x,K)\leq d(x,y)\leq l$.
It follows that $x\in C_{K,l}$.

Claim 4: $C:=\cap_{n}C_{n}$ is compact.
Proof of Claim 4: By Claim 3, $C_{n}$ is closed and hence $C$ is
also closed. Together with Claim 2, it is clear that every sequence
in $C$ has a subsequence that converges to a point in $C$. That
is, $C$ is compact.
A: Recall the Heine-Borel theorem says that a set in a metric space is compact if and only if it is complete and totally bounded. It is easy to see that $C_n$ are closed, so $C$ is closed and therefore complete. It remains to show that $C$ is totally bounded.
Let $\epsilon > 0$ and choose $m \in \mathbb N$ sufficiently large such that $1/ m < \epsilon$. Since $K_m$ is compact, it is totally bounded. Therefore, we can find a finite collection of points $x_k \in X$ such that
$$ K_m \subseteq \bigcup_{k} B(x_k, \epsilon). $$
We claim that
$$ C_m \subseteq \bigcup_k B(x_k, 2\epsilon). $$
Indeed, suppose $x \in C_m$, i.e. $d(x, K_m) \leq 1/m$. By compactness, there exists a witness $y \in K_m$, that is $d(x, y) \leq 1/m$. There is some index $k$ such that $d(x_k, y) < \epsilon$. Thus by the triangle inequality
$$ d(x_k, x) \leq d(x_k, y) + d(x, y) < 2 \epsilon.  $$
Then
$$ \bigcap_n C_n \subseteq C_m \subseteq \bigcup_k B(x_k, 2\epsilon). $$
As $\epsilon$ was arbitrary, this completes the proof. The key ingredient is scaling the radius of the balls $B(x_k, \epsilon)$, which you'll find is a very common argument in the proofs of covering lemmas, which are especially useful in analysis.
A: We can, for free, arrange for $C_n$ to be a decreasing sequence without changing $\cap_n C_n$. Indeed, let (new) $C_n$ be defined as
$$
C_n:=\{x \in X \colon \: d(x,K_j)\leq \frac{1}{j}, \quad \text{for all $j\leq n$.}\}
$$
Then $C_{n} \supset C_{n+1}$ and $\cap_n C_n$ is equal to the original $C$ defined in the OP.
To show that $C$ is compact it suffices to show each $C_n$ is compact. First, let's see why $C_n$ is closed. If $K$ is a subset of $X$, then the function $X \to R$ defined by $x \mapsto d(x,K)$ is continuous -- an easy triangle inequality shows that it is in fact $1$-Lipschitz. Therefore,
$$
X \ni x \mapsto (d(x,K_1),d(x,K_2),\cdots,d(x,K_n)) \in R^n
$$
is continuous. Now, $C_n$ is the pre-image under this map of a closed box in $R^n$.
TBC
