Is $\pi\left(\bigcap_{n}A_n\right)=\bigcap_{n}\pi\left( A_n\right)$, when $\pi$ is a projection?

For me, $\Bbb N$ includes $0$. I am referencing, yet again, this text, exercise $19$, page $30$.

Let $K$ be a compact Hausdorff space, and $\phi:K\to K$ continuous and surjective - i.e. $(K;\phi)$ is a surjective topological dynamic system.

Let $K^\omega=\prod_{n\in\Bbb N}K$, and let $\psi:K^\omega\to K^\omega,\,(x_1,x_2,\cdots)\mapsto(\phi(x_1),x_1,x_2,\cdots)$. By Tychonoff's theorem, $(K^\omega;\psi)$ is a topological system. Let $L=\bigcap_{n\in\Bbb N}\psi^n(K^\omega)\subseteq K^\omega$.

It is "shown" earlier in the book (Corollary $2.27$, page $20$), that $L$ is the maximal (by set inclusion) surjective subsystem of $K^\omega$.

Show that $\pi(L)=K$, where $\pi:K^\omega\to K$ is the projection onto the first component.

I can do this fine, but I fear it is a bit unrigorous in equality $1$:

$$\pi(L)=\pi\left(\bigcap_{n\in\Bbb N}\psi^n(K^\omega)\right)\color{red}{\overset{1}=}\bigcap_{n\in\Bbb N}(\pi\circ\psi^n)(K^\omega)$$Note that $\psi^n(x_1,x_2,\cdots)=(\phi^n(x_1),\phi^{n-1}(x_1),\cdots)$, and $\pi\circ\psi^n$ therefore maps $(x_1,x_2,\cdots)\mapsto\phi^n(x_1)$. As $\phi$ is a surjection on $K$, $(\pi\circ\psi^n)(K^\omega)=K$ regardless of $n$, from which it follows that $\pi(L)=K$.

However, they leave a hint suggesting more rigour is required:

Hint: For $y\in K$ apply Lemma $2.26$ to the $\psi$-invariant set $\pi^{-1}\{y\}$.

Lemma $2.2$6: Suppose that $(K;\phi)$ is a topological system and then $\varnothing\neq A\subseteq K$ is closed and invariant ($\phi(A)\subseteq A$). Then there is a closed set $B$, $\varnothing\neq B\subseteq A$, with $\phi(B)=B$. Explicitly, $B=\bigcap_{n\in\Bbb N_1}\phi^n(A)$.

Assuming for the moment that $\pi^{-1}\{y\}$ is indeed $\psi$-invariant, then this lemma can "solve" the problem similarly (I am unsure why it is needed, but I tried to indulge them nonetheless):

$$\begin{align}\pi(L)&=\pi\left(\bigcap_{n\in\Bbb N}\psi^n(K^\omega)\right)\\&=\pi\left(\bigcap_{n\in\Bbb N}\bigcup_{\mathbf{x}\in K^\omega}\psi^n(\mathbf{x})\right)\\&=\pi\left(\bigcap_{n\in\Bbb N}\bigcup_{y\in K}\psi^n(\pi^{-1}\{y\})\right)\\&\color{red}{\overset{2}{=}}\pi\left(\bigcup_{y\in K}\bigcap_{n\in\Bbb N}\psi^n(\pi^{-1}\{y\})\right)\\&=\pi\left(\bigcup_{y\in K}B_y\right)\\&=\bigcup_{y\in K}\pi(B_y)\\&=\bigcup_{y\in K}y\\&=K\end{align}$$

Why we need to go down that route, I am very unsure. It seems like a strange detour to take, so I feel like I'm missing their intended solution. Moreover, this approach introduces a second dubious equality, $2$, that I don't know how to justify. My proof seems much shorter and more elegant, but also uses a potentially dubious equality in $1$.

Returning to the invariance of $\pi^{-1}\{y\}$ - I do not believe it is invariant:

$$\pi^{-1}\{y\}=\{(y,x_1,x_2,\cdots):x_1,x_2,\cdots\in K\}=\{y\}\times K^\omega\\\psi(\pi^{-1}\{y\})=\{(\phi(y),y,x_1,x_2,\cdots):x_1,x_2,\cdots\in K\}=\{(\phi(y),y)\}\times K^\omega\not\subset\pi^{-1}(y)$$Whenever $\phi(y)\neq y$.

What am I missing with regards to the alleged $\psi$-invariance, and are the equalities $1,2$ correct? That is, is my proposed proof of $\pi(L)=K$ correct?

That $\pi(L)\subseteq K$ is clear.

Here is my take for the rest.

Conditioned on $\pi^{-1}(\left\{y\right\})$ being invariant for all $y\in K$, then (and building on your derivation before identity '2')

$$\pi(L)=\pi\left(\bigcap_{n\in\mathbb{N}}\bigcup_{y\in K} \psi^{n}\left(\pi^{-1}(\left\{y\right\})\right)\right)\supseteq \pi\left(\bigcap_{n\in\mathbb{N}}\bigcup_{y\in K} \psi^{n}\left(C_y\right)\right)=\pi\left(\bigcap_{n\in\mathbb{N}}\bigcup_{y\in K} C_y\right)=\pi\left(\bigcup_{y\in K} C_y\right)=K$$

where in the inclusion above, we resorted to Lemma 2.26 -- in particular, to each $y$ there exists $C_y\subseteq \pi^{-1}(\left\{y\right\})$ with $\psi(C_y)=C_y$.

Regarding the $\psi$-invariance of $\pi^{-1}(\left\{y\right\})$, this would be the case if rather $\psi(x)=\left(x_1,\phi(x_1),x_2,\ldots\right)$ -- which might be blatantly inapropriate for some fundamental reason that I am missing.

This is not true in general. Suppose that $f$ is constant (you always have $f(x)=c$), that each $A_n$ is non-empty and that $\bigcap_{n\in\Bbb N}A_n=\emptyset$. Then$$f\left(\bigcap_{n\in\Bbb N}A_n\right)=\emptyset\quad\text{and}\quad\bigcap_{n\in\Bbb N}f(A_n)=\{c\}.$$