Showing that the stopped filtration $\mathcal{F}_{s \vee t} = \mathcal{F}_s \cup \mathcal{F}_t$

Question

Show that $\mathcal{F}_{s \vee t} = \mathcal{F}_s \cup \mathcal{F}_t$ where $\mathcal{F}_{s}, \mathcal{F}_{t}$ and $\mathcal{F}_{s \vee t}$ are stopping time sigma fields.

Let's suppose that S > T and so $\mathcal{F}_s > \mathcal{F}_t$

$\mathcal{F}_s \cup \mathcal{F}_t \subset \mathcal{F}_s$

This is obvious since, due to the fact that this is an union, $\mathcal{F}_s \cup \mathcal{F}_t$ is $\mathcal{F}_s$ OR $\mathcal{F}_t$ so it's in the case it would be in $F_T$, it will also be in $F_s$.

$\mathcal{F}_s \subset \mathcal{F}_s \cup \mathcal{F}_t$

So:

1. Shouldn't the last be also trivial?

2. In the only case I would have had the intersection of the two sigma algebra instead of the union, the only thing that needs to be shown is that $F_s \in F_t$?