How to show that $\mathbb R$ is not Rothberger, and how to show that it is not Menger? We say that topological space $X$ is Rothberger $S_1(\mathcal O,\mathcal O)$, if: 

For any sequence of open covers of $X$, $\{ \mathcal U_n | n 
\in \omega \}$, one can always find a sequence $\{ U_n | n \in \omega \}$,
  such that for each $n \in \omega$ $\{ U_n \in \mathcal U_n  \}$ and
  $\{ \bigcup_{n \in \omega} U_n \}$ is an open cover of $X$

We say that topological space $X$ is Menger $S_{fin}(\mathcal O,\mathcal O)$, if: 

For any sequence of open covers of $X$, 
  $\{ \mathcal U_n | n \in \omega \}$, 
  one can always find a sequence $\{ A_n | n \in \omega \}$,
  such that for each $n \in \omega$ $\{ A_n \subset \mathcal U_n  \}$, $A_n$ is finite and
  $\{ \bigcup_{n \in \omega} A_n \}$ is an open cover of $X$

I am trying to prove that $\mathbb R$ is not Rothberger and not Menger.
To show that it is not Rothberger. any ideas or directions?
Thank you!
 A: To show that $\mathbb{R}$ does not satisfy the Rothberger property $\mathsf{S}_1 ( \mathcal{O} , \mathcal{O} )$, see my previous answer where I demonstrated that if a subset $A$ of $\mathbb{R}$ has the Rothberger property, then it is strongly measure zero: given any sequence $\langle \varepsilon_n \rangle_{n \in \mathbb{N}}$ of positive real numbers there is a sequence $\langle I_n \rangle_{n \in \mathbb{N}}$ of open intervals such that $A \subseteq \bigcup_n I_n$ and $\mathrm{length} (I_n) \leq \varepsilon_n$ for each $n$.  (Clearly, any $A \subseteq \mathbb{R}$ which is strongly measure zero has Lebesgue measure zero.)
(Avoiding explicit use of this hammer — but still using its key idea — you could just define for each $n \geq 1$ the set $\mathcal{U}_n$ to consist of all open intervals of length $2^{-n}$.  If for each $n \geq 1$ you select some $U_n \in \mathcal{U}_n$, it follows that $\lambda ( \bigcup_n U_n ) \leq \sum_{n=1}^\infty 2^{-n} = 1$, where $\lambda$ denotes the Lebesgue measure. Therefore this union cannot be all of $\mathbb{R}$.)
On the other hand, $\mathbb{R}$ has the Menger property $\mathsf{S}_{\text{fin}} ( \mathcal{O} , \mathcal{O} )$.  This stems from the fact that $\mathbb{R}$ is $\sigma$-compact.

Let $\langle \mathcal{O}_n \rangle_{n \in \mathbb{N} }$ be a sequence of open covers of $\mathbb{R}$.  For each $n$ there is a finite subfamily $\mathcal{A}_n \subseteq \mathcal{U}_n$ which covers the compact interval $[-n,n]$.  It easily follows that $\bigcup_n \mathcal{A}_n$ covers $\mathbb{R}$.

