Measure theory exercise From measure theory volume 1 by Fremlin, exercise 111Xf:
Let $X$ be a set, $\mathcal{A}$ a family of subsets of $X$, and $\Sigma$ the $\sigma$-algebra of subsets of $X$ generated by $\mathcal{A}$. Suppose that
$Y\subset X$. Show that  ${\{E \cap Y : E \in \Sigma}\}$  is the σ-algebra of subsets of $Y$ generated by ${\{A \cap Y : A \in \mathcal{A}}\}$.
I'm able to show that it's a $\sigma$-algebra, and clearly it contains the $\sigma$-algebra generated by ${\{A \cap Y : A \in \mathcal{A}}\}$, but I don't know how to show ${\{E \cap Y : E \in \Sigma}\}$ $\subseteq\sigma (\{{A \cap Y : A \in \mathcal{A}}\})$.

First note that what you want to show is not true in general. You forgot to take the $\sigma$ algebra generated by $\{A\cap Y\mid \dots\}$ instead of just $\{A\cap Y\mid \dots\}$.
To prove the corrected statement, use the good set principle, i.e. set
$$
G = \{E\in \Sigma \mid E\cap Y\in \sigma(\{A\cap Y\mid A\in \mathcal{A}\})\}.
$$
Show that $G$ is a $\sigma$ algebra and that $\mathcal{A}\subset G$.
Then think about why that implies your claim.
