Lemma: If A and B are compact subset of $\mathbb{R}$ then $A\times B$ is compact in $\mathbb{R}^2$.
Proof: let V={$V_{\alpha}\subseteq \mathbb{R}^2 |\alpha \in I$} is an open cover of $A\times B$. So, $A\times B \subseteq \bigcup_{\alpha \in A} V_{\alpha}$. $A$ and $B$ $\subseteq A \times B \subseteq \bigcup_{\alpha \in A} V_{\alpha}$, since $A\times${0} and $B\times${0} are equal to $A$ and $B$. Thus, V is also an open cover of $A$ and $B$. Since $A$ and $B$ are compact, there exist some finite subcover of V i.e. $A\subseteq V_{n_1} \bigcup ..... \bigcup V_{n_k}$ and $B \subseteq V_{m_1} \bigcup ..... \bigcup V_{m_l}$, where $k,l \in \mathbb{N}$. Therefore, $A \times B \subseteq V_{n_1} \bigcup ..... \bigcup V_{n_k} \bigcup V_{m_1} \bigcup ..... \bigcup V_{m_l}$. Which is a finite subcover of V.
Is my proof correct and satisfactory? Can this lemma be generalize to metric space?