Let $I\subset\mathcal{R}$ countable, $(X_t)_{t\in I}$ be a martingale, let $T\in I$ and $\tau$ be a stopping time with $\tau\leq T$.
I do not understand why $E[X_\tau]=E[X_0]$ should hold.
In Probability Theory by A. Klenke (3rd version) the above statement should be just a particular case of the fact that $X_{\tau}=E[X_T|\mathcal{F}_{\tau}$]. This is given in Lemma 10.10, before stating the Optimal Sampling Theorem, by proving that for every $A\in\mathcal{F}_{\tau}$: $$E[X_\tau\mathbb{1}_A]=E[X_T\mathbb{1}_A]$$
Is it sufficient to take $A=\Omega$ and $T=0$? Is the author assuming that $0\in I$?
Yes certainly there is an implicit assumption that $0\in I$ - otherwise there would be absolutely no assumption involving $X_0$ at all!
Similarly the stopping time $\tau$ should take values in $I$.
Once you have $X_\tau=E[X_T|F_\tau]$, you can take expectations on both sides which gives $E X_\tau = E X_T$. (Yes, this is equivalent to taking $A=\Omega$ as you suggest.) I don't think you can directly "take $T=0$" as you said, since that may violate the assumption that $\tau\leq T$. But also you know from the martingale property that $E X_T = E[E[X_T|F_0]]=E[X_0]$ (this assumes that $T\geq 0$; if instead $T<0$ then $E X_0 = E[E[X_0|F_T]]=E[X_T]$).