Q:

Proper map induced from a continuous map

Consider the following claim.
Claim: Let $X$ and $Y$ be locally compact topological spaces, and $f : X \longrightarrow Y$ a continuous map. Suppose that $W \subset X$ is a relatively compact set. Then the induced map (obtained by restricting $f$) $$f : W \backslash f^{-1}(f(\partial W)) \longrightarrow Y \backslash f(\partial W)$$ is a proper map, i.e., the preimage of compact set in $Y \backslash f(\partial W)$ is a compact set in $W \backslash f^{-1}(f(\partial W))$. 
Proof: Let $K \subset Y \backslash f(\partial W)$ be a compact set, we need to show that $f^{-1}(K)$ is a compact set in $W \backslash f^{-1}(f(\partial W))$. By the continuity of $f$, the preimage $f^{-1}(K)$ is closed, so we need only show that $f^{-1}(K)$ does not intersect the boundary of $W \backslash f^{-1}(f(\partial W))$. There are two boundaries  to check: (i) $f^{-1}(K) \cap \partial W = \emptyset$, and (ii) $f^{-1}(K) \cap f^{-1}(f(\partial W)) = \emptyset$. Indeed, since $K \cap f(\partial W) = \emptyset$, (ii) is immediate from the fact that $$f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B),$$ where $A,B$ are some arbitrary sets. Let us therefore turn to (i). Proceeding by contradiction, suppose that $f^{-1}(K) \cap \partial W \neq \emptyset$. Then there exist some $z \in f^{-1}(K) \cap \partial W$, and so $$z \in f^{-1}(K) \cap \partial W \implies f(z) \in f(f^{-1}(K) \cap \partial W) \subset f(f^{-1}(K)) \cap f(\partial W) \subset K \cap f(\partial W) = \emptyset.$$
Concern: I have only (to the best of my knowledge) used the fact that $K$ is a closed subset. The proof is either not correct, or at least not illustrating the main idea, since compactness of $K$ should be necessary. 
Potential Problems: One issue may be specifying exactly what the boundary of $W \backslash f^{-1}(f\partial W))$ is. 

A:

Let $K \subset Y \backslash f(\partial W)$ be a compact set, and let $\{ \mathscr{U}_{\alpha} \}_{\alpha  \in A}$ be an open cover for $f^{-1}(K)$. The set $\mathscr{V} : = \overline{W} \backslash f^{-1}(K)$ is open, so $\{ \mathscr{U}_{\alpha} \}_{\alpha \in A} \cup \mathscr{V}$ is an open cover of $\overline{W}$. Since $\overline{W}$ is compact, it admits a finite sub-cover $\{ \mathscr{U}_{\alpha_1}, ..., \mathscr{U}_{\alpha_n}, \mathscr{V} \}$. This then implies that $\{ \mathscr{U}_{\alpha_1}, ..., \mathscr{U}_{\alpha_n} \}$ is an open cover of $f^{-1}(K)$ which is finite, as required. 

