If $G$ is a sub-filtration on $F$ (i.e.$G_t \subset F_t$), then prove that for a stopping time $\tau$ that $G_\tau \subset F_\tau$ Essentially I want to show that $G_t \subset F_t \Rightarrow G_{\tau} \subset F_{\tau}$. By definition of a stopping time we know $\forall t, \tau \le t$. I'm not too sure how to start on this!
My thinking is using the property that if $\tau$ is a stopping time, then the collection 
$\mathcal{F}_\tau =\{A\in F :A\cap\{\tau \le t\} \in \mathcal{F}_t\}$ would a $\sigma$ algebra, but after this I couldn't get a lead.

It is very much by definition.
So you want to show that $G_\tau \subset F_{\tau}$. Let $A$ be an element of $G_{\tau}$.
For every $t$, by definition $A \cap \{\tau \leq t\} \in G_t$.
Because $G_t \subset F_t$ for each $t$, we know that $A \cap \{\tau \leq t\} \in F_t$.
Since this is true for all $t$, we get that $A \in F_{\tau}$.
Note that $G_t \subset F_t$ for every $t$ is necessary for this proposition to hold. 

In words, an event $A$ in $G_{\tau}$ is one such that for every $t$, if $\tau$ has occurred before $t$ then we can find if $A$ has occurred or not using the information before time $t$. 
If the filtration $G_t$ contains enough information so that we can determine if $\tau$ has occurred before $t$ and then whether $A$ has occurred or not , then the filtration $F_t$ being larger than $G_t$ certainly also contains this information. Hence $A$ would belong in $F_{\tau}$ as well.
