Let $(\Omega,\Sigma)$ be a measurable space and $(F_t)_{t\in I}$ a filtration. We postulate that $F_t$ is the information available at time $t$. Furthermore, if $\tau$ is a stopping time, then $F_\tau$ (the $\sigma$-algebra of the $\tau$-past) is interpreted as the information available at time $\tau$. That being said, I would expect that given some $A\in F_\tau$ and $\omega\in\Omega$ with $\tau(\omega)<\infty$, then $A\in F_{\tau(\omega)}$. Is this correct? I actually even have an informal proof, I hope that someone can comment on it:
Consider an outcome $\omega\in\Omega$ such that $\tau(\omega)<\infty$. Set $t:=\tau(\omega)$, then we know that $\omega\in A$ if and only if $\omega\in A\cap\{\tau\leq t\}$. Since $A\cap\{\tau\leq t\}$ is in $F_t$ (recall that $A\in F_\tau$), we know whether $\omega\in A\cap\{\tau\leq t\}$ and hence we know whether $\omega\in A$. Hence $A\in F_t=F_{\tau(\omega)}$. I feel like I have cheated.
It took me quite some time to figure out where the argument went wrong. As @geetha290krm pointed out in the comment, it is easy to find counter-examples. But, I will try to give some intuition why it should actually not hold.
Saying that $A\in\mathcal{F}_\tau$ means that $A\cap\{\tau \leq t\}\in\mathcal{F}_t$ for any $t$. This can be interpreted by saying that knowing $\{\tau\leq t\}$ and having the knowledge up to time $t$, we can decide whether $A$ happens or not. On the other hand, $A\in \mathcal{F}_t$ means that, with the knowledge up to time $t$ (and nothing additional), we can decide whether $A$ happens or not.
In your argument sou chose some $\omega$ and some $t$ such that $\tau(\omega) = t$. Then, you said that $\omega\in A$ iff $\omega \in A\cap\{\tau \leq t\}$ which is absolutely correct. However, and that is the important points, you do not cover the whole set $A$! More precisely, there might be some $\omega\in A$ such that $\tau(\omega) > t$. For those, your argument doesn't work and that is why the argument breaks down.