real analysis – Help with Baby Rudin Definition 11.4


Hey all, I got stuck on the proof of equation $(12)$. My consideration is as follows:
Assume that $Ain {scr E}$ and $Bin {scr E}$. Then $A=bigcup_{s=1}^n I_s$, $B=bigcup_{k=1}^m J_k$, where $I_s$ and $J_k$ are intervals in $R^p$ for $s=1,2,cdots,n$ and $k=1,2,cdots,m$.
Since $Abigcup B=I_1bigcupcdotsbigcup I_nbigcup J_1bigcupcdotsbigcup J_m$, $Abigcup Bin {scr E}$. It is left to show that $A-Bin {scr E}$.
Note that
A-B&=Abigcap B^c\
&=Abigcapleft(bigcup_{k=1}^m J_kright)^c\
&=Abigcapleft(bigcap_{k=1}^m J_k^cright) (by De Morgan’s Law)\
&=bigcap_{k=1}^mleft(Abigcap J_k^cright)\
&=bigcap_{k=1}^m left( bigcup_{s=1}^n left(I_sbigcap J_k^c right)right)

It can be shown that $bigcap_{k=1}^m left( bigcup_{s=1}^n left(I_sbigcap J_k^c right)right)=bigcup_{s=1}^n left( bigcap_{k=1}^m left(I_sbigcap J_k^c right)right)$, so we only need to show $bigcap_{k=1}^m left(I_sbigcap J_k^c right)$ is an interval.
Now we prove the intersection of any two intervals(namely, $A$ and $B$) is also an interval:
If $Abigcap B=0$, then it is an interval by definition.
If $Abigcap Bneq 0$, then the points $mathbf x=(x_1,x_2,cdots,x_p)$ in $Abigcap B$ has the form
c_ileq x_i leq d_iquad (i=1,2,cdots,p),

so that $Abigcap B$ is an interval.
If we can show that the complement of an interval is also an interval, then the theorem completes. This is where I got stuck. Furthermore, I have no idea about how to show ${scr E}$ is not a $sigma$-ring.
I would appreciate if you could explain in details.