TITLE: Prove that $H$ is compact $\iff$ every cover $\{E_{\alpha}\}_{\alpha \in A}$ has a finite subcovering.
QUESTION [5 upvotes]: Let $H \subseteq \Bbb R^n$. 
Prove that $H$ is compact $\iff$ every cover $\{E_{\alpha}\}_{\alpha \in A}$ where $E_{\alpha}$'s are relatively open in $H$ has a finite subcovering. 

$\bf{Solution \ trial:}$
For $\Rightarrow$ 
Suppose $H$ is compact. 
Suppose $\{E_{\alpha}\}$ are relatively open covering of $H$.
Since $\{E_{\alpha}\}$ are relatively open covering of $H$, 
$\exists$ open set $U_{\alpha}$ such that $U_{\alpha} \cap H= E_{\alpha}$
Then,$\  U_{\alpha}$ is open covering of $H$ 
Since $H$ is compact, $\exists $ finite subset $A_0 \subset A$ such that $$H\subseteq \bigcup_{\alpha \in A_0} \{U_{\alpha}\}$$
Then, $\{E_{\alpha}\}_{\alpha\in A_0} $ is a finite subcovering of  $\{E_{\alpha}\}_{\alpha\in A} $
For $\Leftarrow$
Since  $\{E_{\alpha}\}_{\alpha\in A} $ is relatively open subcovering of H, 
$$\{E_{\alpha}\cap H\}_{\alpha\in A}$$ is relatively open covering. 
$\exists$ a finite subset $A_0 \subset A$ such that $\{ V_{\alpha} \cap H\}_{\alpha \in A_0}$ covers H. 
$$\{ V_{\alpha} \}_{\alpha \in A_0}$$ covers H. 
i.e H is compact.

Is the proof enough? Does There exist any mistake or missings in the detail of the solution? 
Please correct them. Thank you.

REPLY [2 votes]: Assuming the first sentence of the second implication is meant to be ``let $\{V_\alpha\}$ be a covering of $H$ by open subsets of $\mathbf{R}^n$," your argument is correct.