TITLE: Clarification of proof of Theorem 4.14 in Baby Rudin
QUESTION [6 upvotes]: We are proving that $f : X \to Y$, $X$ compact $\Rightarrow f(X)$ is compact.
We reach a step saying 
$$X \subset f^{-1}(V_{\alpha_1}) \cup \dots \cup f^{-1}(V_{\alpha_n})$$
Rudin says that, since $f(f^{-1} (E)) \subset E$, we have
$$f(X) \subset V_{\alpha_1} \cup \dots \cup V_{\alpha_n}$$
Here, is it true that he is implicitly using the following:
First, $A \subset B \to f(A) \subset f(B)$ to say $X \subset f^{-1}(V_{\alpha_1}) \cup \dots \cup f^{-1}(V_{\alpha_n}) \to f(X) \subset f(f^{-1}(V_{\alpha_1}) \cup \dots \cup f^{-1}(V_{\alpha_n}))$
And then saying that $f^{-1}(A) \cup f^{-1}(B) = f^{-1} (A \cup B)$ to say that $f(f^{-1}(V_{\alpha_1}) \cup \dots \cup f^{-1}(V_{\alpha_n})) = f(f^{-1} (V_{\alpha_1} \cup \dots V_{\alpha_n}))$
and then finally using $f(f^{-1} (E)) \subset E$? I'm not sure if there is something less subtle at play here.

REPLY [4 votes]: Yes, that is correct. Alternatively you could say:
$$x\in X\subset f^{-1}(V_{\alpha_1})\cup\cdots\cup f^{-1}(V_{\alpha_n})$$ implies that $x\in f^{-1}(V_{\alpha_i})$ for some $i\in\{1,\dots,n\}$ or equivalently: $$f(x)\in V_{\alpha_i}\subset V_{\alpha_1}\cup\cdots\cup V_{\alpha_n}$$
for some $i\in\{1,\dots,n\}$. This for every $x\in X$ so that: $$f(X)\subset V_{\alpha_1}\cup\cdots\cup V_{\alpha_n}$$