Let $(X_n)_{n\geq0}$ be a stochastic process and $\tau$ a random time. Does the following hold:
$$ \sigma (X_1, \dots, X_\tau) \subset \sigma( X_1, X_2, \dots) \ ?$$
Clearly, $\sigma (X_1, \dots, X_n, \tau ) \not\subset \sigma(X_1,X_2, \dots) $, as the $\tau$ here definitely gives additional information. But what about the above case? Is it obviously true?
Set $\mathcal{F} := \sigma(X_i; i \geq 1)$. If $\tau$ is random time which is $\mathcal{F}$-measurable, then $\{\tau=i\} \in \mathcal{F}$ for all $i$. By definition,
$$X_{\tau} = \sum_{i \geq 1} X_i 1_{\{\tau=i\}}.$$
The finite sum
$$\sum_{i=1}^n X_i 1_{\{\tau=i\}}$$
is $\mathcal{F}$-measurable since $X_i$ and $1_{\{\tau=i\}}$ are $\mathcal{F}$-measurable for each $i$. Hence, the limit
$$X_{\tau} = \lim_{n \to \infty} \sum_{i=1}^n X_i 1_{\{\tau=i\}}$$
is $\mathcal{F}$-measurable (as pointwise limit of $\mathcal{F}$-measurable functions).
This shows that $X_{\tau}$ is $\mathcal{F}$-measurable for any $\mathcal{F}$-measurable random time $\tau$. Hence,
$$\sigma(X_{\tau}; \text{$\tau$ is an $\mathcal{F}$-msb'le random time}) \subseteq \mathcal{F}=\sigma(X_i; i \geq 1).$$ (The converse inclusion is trivial because we can choose $\tau=i$ for fixed $i \geq 1$.)