I am reading Riesz mean ergodic theorem proof and have some questions.

I don't understand where $||p_x-z||<\epsilon$ comes from.I know that if $A\subset R$ and a=infA then for every $\epsilon$>0 i can find another element a' $\in A$ such that $a'<a+\epsilon$ but i don't think this is what he used.Using this i can derive the upper bound of $||z||$ but the author used the fact that $||p_x-z||<\epsilon$ and $||z||=||z-p_x+p_x||$,then apply triangle inequality and get $||z||\leq \mu +\epsilon$.My question is how does one conclude that $||p_x-z||<\epsilon$?
My second question is about the lower of bound of ||z||:Is he using the boundedness of T ? For example $||T^2x||=||T(Tx)||\leq||Tx||\leq ||x||$ and so on for all $T^nx$ .From that we get using the triangle inequality : $||\frac{z+Tz+...+T^nz}{n+1}||\leq ||\frac{n+1||z||}{n+1}||=||z||$ ,is this what he did?
My last question is why the highlighted limit vanishes?

I think that if $T^nx$ converges then it must converge to the fixed point.If $T^nx$ converges to some y then since T is bounded,T is also continuous and so :
$Ty=T(\lim_{n}T^nx)=\lim_{n}T(T^nx)=\lim_{n}T^{n+1}x=y$.Is this what the author implies?
Not so few questions. I will try to answer each of them.
Regarding the last term, this one vanishes because there is the term $\frac{1}{n+1}$ in the limit, which converges to $0$.
Regarding $\Vert p_x -z \Vert< \epsilon$, we can choose such point and such representation of $z$ because $C$ is the closure of a convex set formed by finite linear (positive) combination of the points $\left(T^n x , n \ge 0 \right)$ Concerning the term bounded above by $\Vert z \Vert $, yes, you're right, that's it.
About your last question concerning the idea of author, my answer is "No, he's not meant that". The convergence of $T^n$ needs not to happen everytime, what we can be sure of is the convergence the mean $T^n$. And if we denote $P$ for the mean operator. The author was trying to show that: $TP=P$. I.e $Range(P) $ is a set containing only fixed points of $T$( in this case the vice versa is also true).