Let $Z$ be the simple Galton-Watson Process $Z$ such that $Z_{n}=\sum\limits_{k = 1}^{Z_{n-1}}\xi_{n,i}$ where $(\xi_{n,i})_{n,i\in \mathbb N}$ are IID taking on values in $\mathbb N_{0}$ and $\mathbb E[\xi_{1,1}]=1$ and $P(\xi_{1,1} = 0) > 0$. Note also that $Z_{0}: =1$
Show that $Z_{n} \xrightarrow{n \to \infty} 0$ almost surely $(*)$.
My idea:
I have already checked that this is a martingale so I will not go into those details. By a version of the martingale convergence theorem, we obtain $Z_{n} \xrightarrow{n \to \infty} Z_{\infty}$ almost surely.
Now in order to prove $(*)$, I introduce a stopping time $\tau:=\inf\{n \in \mathbb N: Z_{n} = 0\}$
Clearly, on $\{\tau < \infty\}$ we obtain $Z_{\infty}=0$ almost surely since for $n$ large enough, i.e. $\tau(\omega)\leq n$, we obtain $Z_{m}(\omega)=0$ for all $m \geq n$.
But then consider the event $\{ \tau = \infty\}$,
If $\tau = \infty$, this implies that $Z_{n}\geq 1$ for all $n \in \mathbb N$.
Note that $ \{ Z_{n}\geq 1 \operatorname{ for all }n \}\subseteq \{ \exists i_{n} \leq Z_{n-1} \operatorname{ such that } \xi_{n,i_{n}} > 0 \operatorname{ for all }n \}\subseteq \{\xi_{n,i_{n}} > 0 \operatorname{ for all }n\}\subseteq \{\operatorname{for any n} \exists i :\xi_{n,i} > 0 \}$.
From the independence of $(\xi_{n,i})_{n,i\in \mathbb N}$, we obtain
$P(\operatorname{for any n} \exists i :\xi_{n,i} > 0 \operatorname{ for all }n)=\lim\limits_{n \to \infty}(1-P(\xi_{1,1}=0))^{n}=0$
Hence $P(\tau = \infty)=0$ and thus
$Z_{\infty} = Z_{\infty}\chi_{\tau < \infty} + Z_{\infty} \chi_{\tau = \infty}=Z_{\infty}\chi_{\tau < \infty}= 0$ almost surely.
The key part of my part were the chain of inclusions. Is my proof ok?
This proof is not correct (the statement, though, is correct). The key issue is when you say "from the independence of...": you have chosen the index $i_n$ to depend on $\xi_{n,i_n}$ and so the law of $\xi_{n,i_n}$ has in fact changed. Something in the vein of what you're describing is possible but you'd have to instead union bound over all possible indices $i_n$ for each $n$.
The easiest proof of the fact you seek is through generating functions: if $\phi(s) = \sum_{k} P(\xi = k) s^k$, show that the probability $q$ of death is a fixed point of $\phi$. Use the fact that $1 = \phi'(1)$, $\phi(0) > 0$ and $\phi$ is convex to deduce that the only fixed point in $[0,1]$ is $1$.