As a follow-up to my previous question Show that $T:\,c_0\to c_0\;\;$, $x\mapsto T(x)=(1,x_1,x_2,\cdots),$ is non-expansive. Let $X$ be a normed linear space and $X=c_0$ (the space of sequences of real numbers which converge to $0$). Define \begin{align} T:\,&c_0\to c_0 \\ &x\mapsto T(x)=(1,x_1,x_2,\cdots), \end{align} for arbitrary $x=(1,x_1,x_2,\cdots)\in c_0.$ I want to show that $T$ has no fixed points.
Remark: We have that
$$c_0=\{\bar{x}=(x_1,x_2,\cdots) :x_n\to 0\;\text{as}\;n\to \infty\}.$$
MY TRIAL
Suppose for contradiction that $T$ has fixed points in $X$, then there exists $u\in c_0$ s.t. $T(u)=u.$ That is, \begin{align} T(u_1,u_2,\cdots)=(u_1,u_2,\cdots), \end{align} Please, how do I draw out a contradiction from this?
$(1,u_1,u_2,...)=(u_1,u_2,..)$ implies $1=u_1,u_1=u_2$ etc so $u_n=1$ for all $n$. But then $(u_n) \notin c_0$.