A proof about $F_\sigma$, $\sigma$-compact sets, and subsets of the irrationals I've been looking at a proof that shows the following result. $\mathbb{P}$ is the set of irrational numbers, $\mathbb{Q}$ the rationals, and $\mathbb{R}$ the reals.  

The following conditions are equivalent for any $B \subset \mathbb{P}$
  
  
*
  
*There is a $\sigma$-compact set $S \subseteq \mathbb{P}$ with $B \subseteq S$.
  
*$\mathbb{Q}$ is $G_\delta$ in the subspace $B \cup \mathbb{Q}$ of $\mathbb{R}$.
  

The proof in the text is as follows:
Proof. $(1.) \Leftrightarrow (2.)$. Since a subset of $\mathbb{R}$ is $F_\sigma$ in $\mathbb{R}$ if and only if it's $\sigma$-compact, we have:
\begin{eqnarray*}
(1.)& \Leftrightarrow &B \mbox{ is } F_\sigma \mbox{ in }B \cup \mathbb{Q},\\
 &\Leftrightarrow& S \mbox{ is } F_\sigma \mbox{ in } \mathbb{R} \mbox{ with } S \cap (B \cup \mathbb{Q}) = B,\\
 &\Leftrightarrow& S \mbox{ is } \sigma\mbox{-compact in } \mathbb{R} \mbox{ with }B \subseteq S \mbox{ and } S \cap \mathbb{Q} = \emptyset,\\
&\Leftrightarrow&  (2.). \quad \blacksquare
\end{eqnarray*}
I'm trying to make sense of the implications and clean up the proof using the following result

Proposition 1 Suppose we have a topological space $X$ with $K \subset Y \subset X$. Then, $K$ is compact as a subset of $Y$ if and only if it is compact as a subset of $X$. 

So, this proposition is saying that we can move the compactness property between levels. For the most part, I understand every $\Leftrightarrow$ except for the first one and second ones. 
For the first $\Leftrightarrow$, if we assume $(1.)$, then there is a $\sigma$-compact $S \subseteq \mathbb{P}$ with $B \subseteq S$. Using the proposition, I can show this holds if and only if $B$ is $\sigma$-compact in $\mathbb{R}$ if and only if $B$ is $\sigma$-compact in $B \cup \mathbb{Q}$, but I'm stuck at this point...
As for the second $\Leftrightarrow$, I'm not sure how it would follow. It seems as if it's independent of the first implication.
Attempt
I was also thinking about rearranging the implications as follows.
\begin{eqnarray*}
(1.)& \Leftrightarrow &S \mbox{ is } F_\sigma \mbox{ in } \mathbb{R} \mbox{ with } S \cap (B \cup \mathbb{Q}) = B,\\
&  \Leftrightarrow & B \mbox{ is } F_\sigma \mbox{ in }B \cup \mathbb{Q},\\
 &\Leftrightarrow&  (2.). \end{eqnarray*}
However, the second $\Leftrightarrow$ is throwing me off.
Any suggestions would be greatly appreciated!
 A: I think I figured it out, and it turns out that I was definitely overthinking it. I will post the solution for anyone else that might be interested. All that I was having trouble showing was 
$$ S \mbox{ is }F_\sigma \mbox{ in } \mathbb{R} \mbox{ with } S \cap (B \cup \mathbb{Q}) = B \Leftrightarrow B \mbox{ is }F_\sigma \mbox{ in }B \cup \mathbb{Q}.$$
$(\Rightarrow)$. Suppose $S$ is $F_\sigma$ in $\mathbb{R}$. Then, it can be written as $S = \bigcup_{i=1}^{\infty} K_i$,  where each $K_i$ is closed in $\mathbb{R}$. Since $S \cap (B \cup \mathbb{Q}) = B$, we have $B = \bigcup_{i=1}^{\infty} K_i \cap (B \cup \mathbb{Q})$. Now, since $B \cup \mathbb{Q}$ is a subspace of $\mathbb{R}$, it is equipped witht the relative topology from $\mathbb{R}$, which means that closed sets in $B \cup \mathbb{Q}$ are of the form $K_i \cap (B \cup \mathbb{Q})$, where $K_i$ is closed in $\mathbb{R}$. Therefore, $B$ is $F_\sigma$ in $B \cup \mathbb{Q}$. 
We can revese this to show the $(\Leftarrow)$ implication.
