Inclusion of infinite intersection
No, take $E:=\mathbb{R}$, $x_0:=1$ and $T$ any continuous bounded function with $T(1)=-1$, $T(-1)=1$, $T(0)=2$.
No, take $E:=\mathbb{R}$, $x_0:=1$ and $T$ any continuous bounded function with $T(1)=-1$, $T(-1)=1$, $T(0)=2$.