Closed Map with Compact Fibers is Proper [Follow Up Question]

Question

I've seen the proof provided in this post: Closed map with compact fibers is proper.

It follows the same logic as what is given in the Wikipedia article, and I will adopt the notation used in the Wikipedia proof.

I agree with all of the steps except the very last step, and I'm wondering what it is I'm overlooking. Namely, using the notation from the Wikipedia proof I've linked, I don't agree with:

$$f^{-1}(K) \subset f^{-1}(\cup_{i=1}^{s} \,V_{k_i}) \subset \cup_{\lambda \in \Gamma} \, U_\lambda $$

It seems to me that this assumes:

$$ f^{-1}(Y \setminus f(X \setminus U_{j})) \subset U_{j}$$

And I don't think that is necessarily true unless we assume that $U_{j}$ is saturated (which we do not assume).

Questions:

1. Is the first statement necessarily true without assuming the second statement?
2. If so, why?
3. If not, why would the second statement be true based on the assumptions of the proof?

Answer 1

Small hidden lemma:

if $f: X \to Y$ is a function, and $f^{-1}[\{y\}] \subseteq O$, then $y \in O':=Y\setminus f[X\setminus O]$ and also $f^{-1}[O'] \subseteq O$.

(the first part is mentioned in passing in the Wikipedia proof: "It is easy to check that $V_{k}$ contains the point $k$" because it shows that $\{V_k\mid k \in K\}$ is a cover of $K$. It's an open cover because for a closed map, $O'$ will be open too, etc.).

Proof: if the first were not the case, $y \in f[X\setminus O]$ but then $y=f(x)$ for some $x \in X\setminus O$. Note that $x \in f^{-1}[\{y\}]$ by definition, but $x \notin O$, contradicting $f^{-1}[\{y\}] \subseteq O$! So $y \in Y\setminus f[X\setminus O]$ as required. And if $x \in f^{-1}[O']$, then $x \in O$: otherwise $x \in X\setminus O$, so $f(x) \in f[X\setminus O]$ so $f(x) \notin O'$, but $x \in f^{-1}[O']\ldots $).

In the proof we apply this to $y=k, O= \bigcup_{\lambda \in \gamma_k} U_\lambda$ (the finite subcover that covers $f^{-1}[\{k\}]$) and define $V_k = O'$, for each $k \in K$. So we know that (for each $k$) $f^{-1}[V_k] \subseteq \bigcup_{\lambda \in \gamma_k} U_\lambda$ and this is what allows us to conclude that

$$f^{-1}[\bigcup_{i=1}^k V_{k_i}] \subseteq \bigcup_{\lambda \in \Gamma} U_\lambda$$

because for each $i$ we have $$f^{-1}[V_{k_i}] \subseteq \bigcup_{\lambda \in \gamma_{k_i}} U_\lambda$$ and we collect all the finitely many $\gamma_{k_i}, i=1\ldots s$ into $\Gamma$.