Writing proofs about open covers: Prove continuous function preserves compactness I'd like help improving my proof writing about concepts like open covers or preimages.  These concepts are challenging (at least for me) to write clearly about, because they involve several layers of indirection (subsets of collections of sets, sets of preimages of sets).  Here is a simple proof which I request verification of and improvements to the writing:
Prove that a continuous function on a compact set preserves compactness, using directly the Heine-Borel definition of a compact set as a set for which any open cover admits a finite subcover.
Proof: Let $K$ be a compact set, $f(K)$ its image, and $G = \{g_q : q \in Q\}$ an open cover of $f(K)$.  To show $f(K)$ is compact, we construct a finite subcover $G' \subseteq G$ as follows.
Let $H = \{f^-1(g_q): q \in Q\}$.  Clearly, $H$ is a cover of $K$, and since $f$ is continuous and each $g_q$ open, $H$ is an open cover.  Since $K$ is compact, there exists a finite $Q' \subseteq Q$ such that $\{f^-1(g_q): q \in Q'\}$ covers $K$. Then $G' = \{g_q : q \in Q'\}$ is finite and covers $f(K)$.  Thus, any open cover of $f(K)$ admits a finite subcover, and $f(K)$ is compact.
Discussion: I believe the proof is correct, but not clear or sufficiently explicit.  For example, I state that $G'$ covers $f(K)$ but don't elaborate: any attempt I've made to do so ends up getting tongue-tied talking about individual $g_q$ and the fact that an image of a preimage of a set is that set.  How can the writing be made clear and explicit?
 A: Here is your same exact proof but more explicit:
Let $K \subseteq \mathbb{R}$ be compact and let $f:K \longrightarrow \mathbb{R}$ be continuous. We claim that $f(K) \subseteq \mathbb{R}$ is also compact. Let $C$ be an open cover of $f(K)$. That means that given $y \in f(K)$, there exists an open set $c \in C$ such that $y \in c$.
Step 1: We show that $Q = \{f^{-1}(c): c \in C\}$ is an open cover for $K$. Note that since $f$ is continuous and each $c \in C$ is open, it follows that $f^{-1}(c)$ is open. Let $x \in K$. Then $f(x) \in f(K)$ so there exists $c \in C$ such that $f(x) \in c$. Thus, $x \in f^{-1}(c)$. Since $f^{-1}(c) \in Q$, we get that $Q$ covers $K$.
Step 2: We construct a finite subcover of $f(K)$. Since $K$ is compact, we may choose $q_{1}, \dots, q_{n}$ to be a elements of $Q$ that cover $K$. Since each $q_{i}$ belongs to $Q$, there exists a set $c_{i} \in C$ such that $f^{-1}(c_{i}) = q_{i}$ (by definition of $Q$). Let $c_{1}, \dots, c_{n}$ be elements of $C$ such that $f^{-1}(c_{i}) = q_{i}$.
Step 3: We show that the sets $c_{1}, \dots, c_{n}$ form an open cover of $f(K)$. Let $y \in f(K)$. Then $y = f(x)$ for some $x \in K$. Take $q_{i}$ such that $x \in q_{i}$. Recall that this is possible because the sets $q_{1}, \dots, q_{n}$ cover $K$. Then $x \in f^{-1}(c_{i})$ (by definition of $c_{i}$). Hence, $y = f(x) \in c_{i}$. This completes the proof.
