Let's say $A=\{[-a,2a]\mid a\in[0,\infty)\}$. I want to prove that it doesn't generate the $\sigma$-algebra on $\mathbb{R}$.
I have no idea how to start this prove because I thought it would produce the $\sigma$-algebra on $\mathbb{R}$.
Can someone explain why it's not and how to start a prove like this.
Assuming you meant what @WoolierThanThou commented:
Your generating sets have a certain sort of "symmetry", using the term "symmetry" a bit loosely.
Find a clever bijective map $f: \mathbb R \rightarrow \mathbb R$ that expresses that symmetry, i.e. preserves every element of $A$, ($f([-a,2a]) = [-a, 2a]$ for $a \ge 0$), but doesn't preserve many others.
Show that this property extends to the $\sigma$-algebra generated by $A$.
Show that there is at least one set in the standard Borel $\sigma$-algebra that isn't preserved by $f$.
If you're having trouble coming up with an appropriate $f$,
Update (10 Jan 2024) Having done most of what was outlined above on a different post, I figured I should update this one with the details and a link to the other post.
Consider the map $\beta: \mathbb R \rightarrow \mathbb R$ given by
$$ \beta(x) = \begin{cases} -2x, & \text{if $x \lt 0$} \\ -x/2, & \text{if $x \ge 0$} \end{cases}$$
$\beta$ behaves like reflecting the real number line through the point $x=0$, except it also stretches the negative half and smushes the positive half. The important things about $\beta$ are
1 - $\beta( [-2a,a] ) = [-2a, a]$, i.e. all the sets in $A$ are invariant under $\beta$, and
2 - $\beta \circ \beta = id_{\mathbb R}$, i.e. it act like a symmetric reflection in that doing it twice brings you back to your original state.
Now we can use the fact that "the symmetry under $\beta$ extends to the generated $\sigma$ algebra" (proved here), and get the result that all the sets in the $\sigma$ algebra generated by $A$ are invariant under $\beta$ as well. As a specific example, for the Borel set $[-2,2]$, $\beta( [-2,2] ) = [-1,4]$, and is thus not in $\sigma(A)$.