Rudin 2.10 (b) Example

Note that $\bigcap_{x\in A}E_x\subset(0,\infty)$. If $y\in \bigcap_{x\in A}E_x\subset(0,\infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).

Regarding your comment, the nested interval property is about compact sets. These are open and not closed.

Note that $x\notin E_x$ for every $x$