In Tom Apostol’s mathematical analysis at page 82, I am confused about the proof of theorem 4.25
Theorem 4.25. Let $f : S \to T$ be a function from one metric space $(S, d_S)$ to another $(T, d_T)$. If $f$ is continuous non a compact subset $X$ of $S$, then the image $f(X)$ is a compact subset of $T$; in particular, $f(X)$ is closed and bounded in $T$.
Proof. Let $F$ be an open covering of $f(X)$, so that $f(X) \subseteq \bigcup_{A \in F} A$. We wil show that a finite number of the sets $A$ cover $f(X)$. Since $f$ is continuous on the metric subspace $(X, d_S)$ we can apply Theorem 4.23 to conclude that each set $f^{-1}(A)$ is open in $(X, d_S)$. The sets $f^{-1}(A)$ form an open covering of $X$ and, since $X$ is compact, a finite number of them cover $X$, say $$X \subseteq f^{-1}(A_1) \cup \cdots \cup f^{-1}(A_p).$$ Hence \begin{align*} f(X) &\subseteq f[f^{-1}(A_1) \cup \cdots \cup f^{-1}(A_p)] \\ &= f[f^{-1}(A_1)] \cup \cdots \cup f[f^{-1}(A_p)] \\ &\subseteq A_1 \cup \cdots \cup A_p, \end{align*} so $f(X)$ is compact. As a corollary of Theorem 3.38, we see that $f(X)$ is closed and bounded.
It is said that $f^{-1}(A)$ is open in $(X,d_S)$, yes, but the sets $f^{-1}(A)$ are not an open covering of $X$, because an open covering of $X$ is a collection $F$ of open subsets of $S$ and $X$ belongs to these open sets
$X$ is subspace of $S$, $f^{-1}(A)$ is open in $X$, but it not always open in S
If the theorem said that $f$ is continuous on $S$, I think the proof is ok, but it said $f$ is continuous just on $X$.
I am confused, please help me, where have I made a mistake?
Let $x \in X$ and $y=f(x)$. Then $y \in f(X)$ which implies $y$ is in one of the open sets $A$. This implies $x \in f^{-1}(A)$, so $X \subset \cup_A f^{-1}(A)$. This is what is meant by saying that sets $f^{-1}(A)$ form a cover for $X$.