Proof that the nested descreasing intersection of nonempty covering compact sets is nonempty.
Let $$A_{1} \supset A_{2} \supset \cdots \supset A_{n} \supset \cdots$$ be a nested decreasing sequence of compacts. Suppose that $\displaystyle \bigcap A_{n} = \emptyset$. Take $U_{n} = A_{n}^{c}$, then $$\bigcup U_{n} = \bigcup A_{n}^{c} = \left(\bigcap A_{n}\right)^{c} = A_{1}.$$ Here, I'm thinking of $A_{1}$ as the main metric space. Since $\{U_{n}\}$ is an open covering of $A_{1}$, we can extract a finite subcovering, that is, $$A_{\alpha_{1}}^{c}\cup A_{\alpha_{2}}^{c} \cup \cdots \cup A_{\alpha_{m}}^{c} \supset A_{1}$$ or $$(A_{1}\setminus A_{\alpha_{1}})\cup(A_{1}\setminus A_{\alpha_{2}})\cup\cdots\cup(A_{1}\setminus A_{\alpha_{m}}) \supset A_{1}.$$ But, this is true only if $A_{\alpha_{i}} = \emptyset$ for some $i$, a contradiction.
Is correct?
In my opinion, you should add a justification of the assertion according to which that$$(A_1\setminus A_{\alpha_1})\cup(A_1\setminus A_{\alpha_2})\cup\cdots\cup(A_1\setminus A_{\alpha_m})\supset A_1\tag1$$can only occur if $A_1=\emptyset$. This happens because$$(1)\iff A_{\alpha_1}\cap\cdots\cap A_{\alpha_m}=\emptyset,$$but$$A_{\alpha_1}\cap\cdots\cap A_{\alpha_m}=A_{\max\{\alpha_1,\ldots,\alpha_m\}}\neq\emptyset.$$