Show that $\mathbb R$ is countably compact 
Show that $\mathbb R$ is countably compact, that is that any cover has an at most countable subcover.

I tried to use the fact that every open set in R is a union of intervals with rational endpoints.
I tried to use the below fact first.
[If every open subset of $\mathbb R$ is a disjoint union of open intervals, the number of the intervals is at most countable] 
but I realized that this is just a special case, 
and what I need to prove is that "any" cover has at most countable subcovers.
Could you help me please?
Thank you!
 A: To repeat what was noted in the comments, you’re not showing that $\Bbb R$ is countably compact: you’re showing that it’s Lindelöf. The difference between the two properties lies largely in the placement of the word countable:

A space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. It’s Lindelöf if every open cover of $X$ has a countable subcover.

Your idea of using the fact that every open set in $\Bbb R$ is a union of open intervals with rational endpoints is a good one. It works: you just have to fill in some details. Let $\mathscr{B}$ be the set of open intervals with rational endpoints, and let $\mathscr{U}$ be any open cover of $\Bbb R$. For each $x\in\Bbb R$ choose a $U_x\in\mathscr{U}$ and a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq U_x$. (How do you know that this is possible?) Now let $\mathscr{B}_0=\{B_x:x\in\Bbb R\}$.


*

*Explain why $\mathscr{B}_0$ is a countable open cover of $X$.  

*For each $B\in\mathscr{B}_0$ there is a $U_B\in\mathscr{U}$ such that $B\subseteq U_B$; why?  

*Show that $\{U_B:B\in\mathscr{B}\}$ is a countable subfamily of $\mathscr{U}$ that covers $\Bbb R$.  

*Since $\mathscr{U}$ was an arbitrary open cover of $\Bbb R$, conclude that $\Bbb R$ is Lindelöf.

