$\gdef\In{\subset}$
The theorem states that, for the metric space $\R^n$, a subset is compact if and only if it is closed and bounded.
We will prove this theorem for $\R$ first, then we show that if $A \In X$ is compact and $B \In Y$ is compact, then $A \times B \In X \times Y$ is compact.
For any metric space $X$, a compact subset $K$ is bounded, i.e there exists a real number $M>0$, such that for any $x,y \in K$, $d(x,y) < M$. Suppose not, then for any $x_0 \in K$, and $n > 0$, $K \RM B_n(x_0)$ is not empty. Consider the open cover $\cup_{n>0} B_n(x_0)$, then it has no finite sub-cover.
Hence, we have proven that, compactness implies closed and bounded.
The converse in general is not true. Consider any infinite set $X$ with discrete metric, $d(x,y) = \delta_{x,y}$. Then, $X$ is closed and bounded, but not compact, since $X = \cup_{x \in X} \{x\}$, and $\{x\} = B_{1/2}(x)$ are open sets.
It is true in $\R, \R^n$.
Proposition Any closed interval $[a,b]$ is compact. Let $\{U_\alpha\}$ be an open cover of $[a,b]$. Suppose there is no finite subcover, then we can divide the interval in half and find out which half is causing the trouble. Let $I_0 = [a,b]$, let $I_1 \In [a,b]$ be either $[a, (a+b)/2]$ or $[(a+b)/2, b]$, that does not admit finite subcover. Repeating this process, we get a nested sequence of closed intervals $I_0 \supset I_1 …$, the diameter $|I_n| = 1/2 |I_{n-1}|$. Let $I_n =[a_n,b_n]$, then $a_n$ is a bounded increasing sequence, and $b_n$ is a bounded decreasing sequence. We have a limit $\cap_n I_n = \{x\}$. But $\{x\}$ is counted in certain $U_\alpha$, and there is a $r>0$, such that $B_r(x) \In U_\alpha$. Hence, if $|I_n| < r$, we have $I_n \In B_r(x) \In U_\alpha$, contradicting with $I_n$ does not admit finite subcover.
Corollary Any closed bounded subset in $\R$ is compact. Since it is a closed subset, and contained in a compact subset $[-M, M]$ for $M$ large enough.
Proposition If $\{K_\alpha\}$ is a collection of compact sets, such that any finite subcollection of them has non-empty intersection, then $\cap_{\alpha} K_\alpha \neq \emptyset$.
Proof: Suppose not, then $\cup_{\alpha} K_\alpha^c = X$. Choose any $\alpha_0$, then $\cup_{\alpha \neq \alpha_0} K_\alpha^c \supset K_{\alpha_0}$. By compactness, we have a finite sub-cover of $K_{\alpha_0}$ $$ K_{\alpha_0} \In K_{\alpha_1}^c \cup \cdots \cup K_{\alpha_n}^c $$ This means $$ K_{\alpha_0} \cap K_{\alpha_1} \cap \cdot \cap K_{\alpha_n} = \emptyset $$ get contradiction.
Proposition If $E \In K$ is an infinite subset, then $E$ has a limit point in $K$.
If not, then every point in $x \in K$ has an open ball $U_x$ containing at most one point in $E$. By compactness of $K$, $K$ is covered by finitely many such $U_x$, hence $E$ is also covered by such finitely many $U_x$, which contradicts with $E$ being infinite.
Corollary If $(x_n)$ in $K$ is a sequence, then there exists a convergent subsequence. (Compactness implies sequential compactness.)
If the value set $V = \{x_n\}$ is finite, then there is a value which appears infinite many times in the sequence, and we can take the constant subsequence. If the value set is infinite, then we by proposition, there exists a $x \in K$, such that for any $r>0$, $B_r(x) \cap V$ is an infinite set, hence $\{n \in \N| x_n \in B_r(x) \}$ is infinite, thus $x$ is a subsequential limit of $(x_n)$.