Let $\sigma$ be a stopping time and $Z$ a $\mathcal{F}_{\sigma}$-measurable random variable. Now, I want to show that for any $A \in \mathcal{B}_{[0,t]}$, $\mathbb{1}_{\{\sigma \in A\}} Z$ is $\mathcal{F}_t$-measurable.
I want to start with an interval, say $A=[0,s]$, since this interval generates a borel $\sigma$-algebra. But for me it is unclear how I can work towards the application of the $\pi$-$\lambda$-theorem to prove the above statement.
Since $Z$ is $\mathcal{F}_{\sigma}$-measurable, we know that $\{Z\in B\}\cap \{\sigma\leq t\}\in \mathcal{F}_t$ for all $t$ and all Borel sets $B$. Another way of writing this is that $1_{\sigma\leq t}Z$ is $\mathcal{F}_t$-measurable for all $t$.
In your problem, $1_{\sigma\in A}=1_{\sigma\in A}1_{\sigma\leq t}$ since $A$ is a subset of $[0,t]$. Therefore $$1_{\sigma\in A}Z=1_{\sigma\in A}1_{\sigma\leq t}Z=1_{\sigma\in A}(1_{\sigma\leq t}Z) $$ so it's enough to show that $1_{\sigma\in A}$ is $\mathcal{F}_t$-measurable.
Let $\mathcal{S}$ be collection of all Borel subsets $A$ of $[0,t]$ for which $1_{\sigma\in A}$ is $\mathcal{F}_t$-measurable. As you noted, $\mathcal{S}$ contains all intervals $[0,s]$ with $s\leq t$, which is a $\pi$-system generating the Borel $\sigma$-algebra on $[0,t]$. So it suffices to show that $\mathcal{S}$ is a $\lambda$-system.
$\mathcal{S}$ contains $[0,t]$, and if $A,B\in\mathcal{S}$ with $A\subset B$, then $1_{\sigma\in B\setminus A}=1_{\sigma\in B}-1_{\sigma\in A}$ is also $\mathcal{F}_t$ measurable. Finally, if $\{A_n\}$ is a sequence in $\mathcal{S}$ increasing to $A$, then $1_{\sigma\in A}=\lim_{n\to\infty}1_{\sigma\in A_n}$, hence $1_{\sigma\in A}$ is also $\mathcal{F}_t$ measurable. This proves that $\mathcal{S}$ is a $\lambda$-system.