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