We have following 3 definitions.
Definition: Suppose that $(X,d)$ is a metric space. A sequence $(\textbf{x}_n)_{n\in \mathbb{N}}$ of points in $X$ is said to be a Cauchy sequence, if, given any $\epsilon>0$, there exists $N$ such that $d(\textbf{x}_m,\textbf{x}_n)<\epsilon$ whenever $N\le n<m$.
Definition: Suppose that $(X,d)$ is a metric space, and that $S\subseteq X$. We say that $S$ is complete if and only if every Cauchy sequence in $S$ has a limit in $S$.
Definition: Suppose that $(X,d)$ is a metric space, and that $S\subseteq X$. We say that $S$ is compact if and only if every sequence in $S$ has a convergent subsequence with limit in S.
And are the proofs of the 3 following theorem correct?
Theorem 1: Suppose that $(X,d)$ is a metric space. Intersection of any collection of complete subsets of $X$ is complete.
Proof: Let $(\textbf{x}_n)$ be a Cauchy sequence in $\bigcap_{i\in I}A_i$, an intersection of any collection $\{A_i\}_{i\in I}$ of complete subsets of $X$. Since $(\textbf{x}_n)$ is a Cauchy sequence in $\bigcap_{i\in I}A_i$, it is a Cauchy sequence in $A_1$. Since $A_1$ is complete, $\textbf x_n \rightarrow \textbf x$ where $\textbf x \in A_1$. Similarly, since $(\textbf{x}_n)$ is a Cauchy sequence in $\bigcap_{i\in I}A_i$, it is a Cauchy sequence in $A_2$. Since $A_2$ is complete, $\textbf x \in A_2$. Without loss of generality, $\textbf x \in A_i$ for all $I\in I$.
$\therefore x\in \bigcap_{i\in I}A_i$. Hence $\bigcap_{i\in I}A_i$, the intersection of any collection of complete subsets of $X$ is complete.
Theorem 2: Suppose that $(X,d)$ is a metric space. The intersection of a compact set and a closed set in $X$ is compact.
Proof: Denote $E\subseteq X$ to be the compact set in $X$ and $W$ to be a closed set in $X$.
We know that a compact set in a metric space is also complete and complete sets are closed in $X$. Therefore $E$ is closed. Now, $E \cap W$ is a closed set, since intersection of any collection of closed sets is closed. And $E \cap W \subseteq E$.
Now, suppose that $(\textbf{x}_n)$ be a sequence in $E \cap W$ and so in $E$. Since $E$ is compact, $(\textbf{x}_n)$ has a convergent subsequence with limit $\textbf x$, say. If $\textbf x=\textbf x_n$ for some $n \in \mathbb{N}$, then clearly $\textbf x \in E\cap W$. If $\textbf x \neq \textbf x_n$ for any $n\in \mathbb{N}$, then $\textbf x$ is an accumulation point of $E\cap W$, so that $\textbf x \in E\cap W$ since $E\cap W$ is closed.
Hence, $E\cap W$, the intersection of a compact set and a closed set in $X$ is compact, since every sequence in $E\cap W$ has a convergent subsequence in $E\cap W$.
Theorem 3: Suppose that $(X,d)$ is a metric space. The intersection of any collection of compact sets in $X$ is compact.
Proof: Let $\{A_i\}_{i\in I}$ be any collection compact sets of $X$. For all $I\in I$, $A_i$ is a closed set, since compact sets are closed. Moreover, intersection of any collection of closed sets in $X$ is closed, hence $\bigcap_{i\in I}A_i$ is a closed set. Also, we have $\bigcap_{i\in I}A_i \subseteq A_1$.
Now, suppose that $(\textbf{x}_n)$ be a sequence in $\bigcap_{i\in I}A_i$ and so in $A_1$. Since $A_1$ is compact, $(\textbf{x}_n)$ has a convergent subsequence with limit $\textbf x$, say. If $\textbf x=\textbf x_n$ for some $n \in \mathbb{N}$, then clearly $\textbf x \in \bigcap_{i\in I}A_i$. If $\textbf x \neq \textbf x_n$ for any $n\in \mathbb{N}$, then $\textbf x$ is an accumulation point of $\bigcap_{i\in I}A_i$, so that $\textbf x \in \bigcap_{i\in I}A_i$ since $\bigcap_{i\in I}A_i$ is closed.
Hence, $\bigcap_{i\in I}A_i$, the intersection of any collection of compact sets in $X$ is compact, since every sequence in $\bigcap_{i\in I}A_i$ has a convergent subsequence in $\bigcap_{i\in I}A_i$.
And how would you prove the following theorem?
Theorem: Suppose that $(X,d)$ is a metric space. Union of a finite number of complete subsets of $X$ is complete. ($(X,d)$ may or may not be complete)
I find proving this last theorem ok if $X$ was complete, but when we don't know whether $X$ itself is a complete metric space or not, I found that it is hard to prove this theorem.
Best Answer
Your proofs seem fine at first glance, except you need not assume that your index sets are countable.
As for your question : Let $S = S_1\cup S_2\cup \ldots S_n$ be the union of a finite number of complete subsets of $X$. Choose a Cauchy sequence $(x_n) \in S$, then there is some $k \in \{1,2,\ldots, n\}$ such that infinitely many $x_n$'s are in $S_k$ (otherwise the entire sequence would be a finite set, which is the trivial case we can ignore).
Thus, there is a subsequence $(x_{n_j}) \in S_k$ which must have a limit in $S_k$ (since $S_k$ is complete). However, if a subsequence of a Cauchy sequence converges to a point, then the entire sequence converges to that point. (Why?)
Hence that point is in $S$