Concrete Filtration at specific time Consider random variables $X(t) \geq 0$ for $t=0,1,2,3$ defined on a filtered  probability space with $\Omega=\{u,d\}^3$, where $u>d$ and $u,d \geq 0$.
I holds  $$X(t)(\omega)=w_t$$ for any $\omega=(\omega_1,.., \omega_3)$
The filtration at $t$ is generated by $X(t)$, meaning : $\mathcal{F}(t)=\sigma(X(1),..,X(t))$
For $t=0$ define $\mathcal{F}(0)=\{ \emptyset, \Omega\}$.
I want to compute $\mathcal{F}(1),\mathcal{F}(2)$.
Since I know at $t=1$, which state $X(1)$ has taken, I would conclude: $$\mathcal{F}(1)=\{ (u,u,u), (u,d,d), (u,u,d), (u,d,u)\}$$
I assumed $X(1)=u$.
In similar way I would argue for $t=2$:$$ \mathcal{F}(2)=\{ (u,u,d), (u,u,d)\}$$
assuming $X(1)=X(2)=u$.
What do you think?
 A: The meaning of a filtration
See, the idea of a filtration is of storing accumulated information.
What does this mean? Well, imagine you are travelling on a bus, on a route you've never been on. You note down the stops that come your way : when you are at a certain time, you know everything about a the place you are at and everything before it, but you don't quite know what comes afterwards. That's the point of a filtration : to store the list of events that you know have happened within a certain time, or are decided by a certain list of random variables, you use a filtration.
For example  by time $1$ you are expected to have seen $X_1$, and therefore $\mathcal F(1)$ is the list of events that you know for sure have occurred or not, by only looking at $X_1$.
Note that an event is a subset of the sample space i.e. $\Omega$. So $\mathcal F(1)$ is a subset of $\mathcal P(\Omega)$, the set of subsets of $\Omega$. Every element of $\mathcal F(1)$ is a subset of $\Omega$. You seem to have written just one event when you wrote down $\mathcal F(1)$ ; that event turns out to be a good example though.
One can similarly speak of $\mathcal F(2)$.
Key point : The occurence of an event in $\mathcal F(1)$ depends $\color{green}{\text{only on } X_1}$, and every such event will belong to $\mathcal F(1)$.
Why do you think that $\mathcal F(0) = \{\emptyset , \Omega\}$? This is because these are the two events you can be certain have occured or not without knowing anything : for any other event, you will have to know something about the process to know if it occurred or not.
Examples
We will take examples, and in the process point out a stunning commonality of the affirmative cases, with the help of which I will mention a general pattern that one can use to form the filtrations in the blink of an eye.
Let's take the event $E_1 = \{(u,u,d)\}$. Is this event in $\mathcal F(1)$? Can I decide using just $X_1$ if $E_1$ has occurred or not? The answer is no, because if $X_1 = u$ occurs, I'd still have to check if $X_2 = u$ and $X_3 = d$ happen, only then can $E$ happen. So I need to go beyond $X_1$ to check if $E_1$ occurs or not.
Let's take $E_2 = \{(u,u,d),(u,d,u),(d,d,u)\}$. Is this event in $\mathcal F(1)$? Once again, the answer is no : for example, if $X_1 = u$ then I can't conclude that $E_2$ has occurred : what if $X_2= X_3 = d$ happens? Then $E_2$ won't happen. So $E_2$ very much depends upon $X_2,X_3$.
Let's take $E_3 = \{(u,u,d),(u,d,u),(u,d,d),(u,u,u)\}$. Is this event in $\mathcal F(1)$? The answer is YES. Here's why : if $X_1 = u$ then regardless of what $X_2$ and $X_3$ are, the event $E_3$ occurs. On the other hand, if $X_1 = d$ then once again the event certainly does not occur. In conclusion, $E_3$ depends only on $X_1$.
Let's take $E_4 = \{(u,u,d),(u,d,u),(u,d,d),(u,u,u) , (d,u,u)\}$. Is this event in $\mathcal F(1)$? The answer is NO, because if $X_1 = d$ then it is still possible that $X_2 = d$ so then $E_4$ won't occur.
Now we are beginning to see things : what's happened in $E_3$ is that after $X_1 = u$, all possible combinations of $X_2$,$X_3$ were covered in $E_3$. Because of that , the event $E_3$ said : I don't care about $X_2$ or $X_3$, just tell me if $X_1 = u$ or not. The other events did not . $E_4$ said :  if $X_1 = u$ then I occur, but if $X_1 =d$ then you'll need to check $X_2$ and $X_3$  to see if I occur or not.
This idea of covering all possible future combinations is the key to forming sets that belong to filtrations. Including all possible combinations says "I don't care about those coordinates" and that's what you want, isn't it?
Observing the pattern
The pattern is thus noticed as follows : if you want an event to depend only on a certain bunch of random variables, make sure that the event includes all possible combinations of the rest of the variables.
Now, what does that mean in the context of $\mathcal F(1) = \sigma(X_1)$? Well, you need all possible combinations of $X_2,X_3$ in any event in $\mathcal F(1)$.
Of course both the empty set and $\Omega$ belong to $\mathcal F(1)$. There are only two other possible events in $\mathcal F(1)$. Why? Well, let's imagine an event $E \in \mathcal F(1)$. If it contains $(u,*,*)$ for any value of $*$ then it must contain all possible combinations so that the value of those $*$s don't matter for $E$. In particular, it must contain $E_3 = \{(u,u,d),(u,d,u),(u,u,u),(u,d,d)\}$.
Similarly, if $E$ contains $\{(d,*,*)\}$ for any value of $*$ then it must contain all possible combinations ,so it must contain $\{(d,u,u),(d,u,d),(d,d,u),(d,d,d)\}$.
That gives us only these sets, which form $\mathcal F(1)$ :
$$
\mathcal F(1) =\Big\{\emptyset , \Omega , \{(u,u,d),(u,d,u),(u,u,u),(u,d,d)\} , \{(d,u,u),(d,u,d),(d,d,u),(d,d,d)\} \Big\}
$$
If you take any other set, it will depend on $X_1$. Try it!
A different way of understanding $\mathcal F(1)$
That is through the following lemma :

$\mathcal F(1)$ is precisely the set of all events that can be obtained as preimages of $X_1: \Omega \to \mathbb R$.

So what we need to do, is characterize the preimages of $X_1$. Let $S \subset \mathbb R$. Then $\{\omega : X_1(\omega) \in S\} = \{\omega : \omega_1 \in S\}$. But then, this event doesn't depend upon $\omega_2$ and $\omega_3$.  Now see the magic :
$$
\emptyset =  \{\omega : X_1(\omega) \in \emptyset\}\\
\Omega =  \{\omega : X_1(\omega) \in \{u,d\}\} \\
\{(u,u,d),(u,d,u),(u,u,u),(u,d,d)\}  = \{\omega : X_1(\omega) \in \{u\}\} \\
\{(d,u,u),(d,u,d),(d,d,u),(d,d,d)\} = \{\omega : X_1(\omega) \in \{d\}\}
$$
now do you understand the beauty of the lemma? We obtain the sets we need as preimages of $X_1$ under certain well chosen sets.
Onto $\mathcal{F}(2)$
For $\mathcal F(2)$ we again use free choice for the last coordinate $X_3$. So if an event in $\mathcal F(2)$ contains $(u,d,*)$ it must contain all values of $*$ so that the third coordinate stops mattering for that event.
With that, we get four fundamental sets , corresponding to the first two combinations of $X_1,X_2$ as $(u,u,*),(u,d,*),(d,u,*)$ and $(d,d,*)$:
$$
F_1 = \{(u,u,d),(u,u,u)\} \\
F_2 = \{(u,d,d),(u,d,d)\} \\
F_3 = \{(d,u,d),(d,u,u)\} \\
F_4 = \{(d,d,d),(d,d,u)\}
$$
Any event in $\mathcal F(2)$ is a union of the events $F_1,F_2,F_3,F_4$. For example $F_1 \cup F_2 \in \mathcal F(2)$, $F_3 \cup F_4 \cup F(1) \in \mathcal F(2)$. So $\mathcal F(2)$ is precisely given by the set $$\mathcal F(2) = \{\cup_{i \in \mathcal I} F_i : \mathcal I \subset \{1,2,3,4\}\}$$
(Note : the empty union gives $\emptyset$ and $\Omega = F_1 \cup F_2 \cup F_3 \cup F_4$, so they very much belong to the filtration).
And how to get this using the preimage method? Some examples are given from which I'd assume you'd be able to get the general idea :
$$
F_1 = \{\omega : (X_1(\omega),X_2(\omega)) \in \{(u,u)\}\} \\
F_3 = \{\omega : (X_1(\omega),X_2(\omega)) \in \{(d,u)\}\} \\
F_2 \cup F_4 = \{\omega : (X_1(\omega),X_2(\omega)) \in \{(u,d), (d,d)\}\}
F_1 \cup F_2 \cup F_4 = \{\omega : (X_1(\omega),X_2(\omega)) \in \{(u,u), (u,d), (d,d)\}\}
$$
Try to see that $\mathcal F(1) \subset \mathcal F(2)$ (hint : try to write each element of $\mathcal F(1)$ as a union of the $F_i$).
