Here is Theorem 29.1 in the book Topology by James R. Munkres, 2nd edition:
Let $X$ be a [topological] space. Then $X$ is locally compact Hausdorff if and only if there exists a [topological] space $Y$ satisfying the following conditions:
(1) $X$ is a subspace of $Y$.
(2) The set $Y - X$ consists of a single point.
(3) $Y$ is a compact Hausdorff space.
If $Y$ and $Y^\prime$ are two spaces satisfying these conditions, then there exists a homeomorphism of $Y$ with $Y^\prime$ that equals the identity map on $X$.
Here is the paragraph in Step 2 of the proof showing that the space $Y$ is compact:
To show that $Y$ is compact, let $\mathscr{A}$ be an open covering of $Y$. The collection $\mathscr{A}$ must contain an open set of type (2) say [a set of the form] $Y - C$, [where $C$ is a compact subspace of $X$] since none of the open sets of type (1) [which are open sets of $X$] contains the point $\infty$ [which is the only point of the singleton set $Y - X$]. Take all the members of $\mathscr{A}$ different from $Y-C$ and intersect them with $X$; they form a collection of open sets of $X$ covering $C$. Because $C$ is compact, finitely many of them cover $C$; the corresponding finite collection of elements of $\mathscr{A}$ will, along with the element $Y - C$, cover all of $Y$.
Now how about the following modification (perhaps even a simplification!) of this part of the proof?
To show that $Y$ is compact, let $\mathscr{A}$ be an open covering of $Y$. The collection $\mathscr{A}$ must contain an open set of type (2) say [a set of the form] $Y - C$, [where $C$ is a compact subspace of $X$] since none of the open sets of type (1) [which are open sets of $X$] contains the point $\infty$ [which is the only point of the singleton set $Y - X$]. Take all the members of $\mathscr{A}$ different from $Y-C$; they form a collection of open sets of $Y$ covering $C$. Because $C$ is compact [as a subspace of $X$ and since $X$ in turn is a subspace of $Y$, this set $C$ is also compact as a subspace of $Y$], finitely many of these sets cover $C$; these finitely many sets of $\mathscr{A}$ will, along with the set $Y - C$, cover all of $Y$.
Is this argument a valid one? Or, are there any problems in it? If valid, then isn't this approach a little simpler than that adopted by Munkres?
The only difference, it seems to me, is that Munkres intersects all the sets in $\mathcal{A}$ with $X$, and indeed this is not strictly needed, as compactness is "absolute" (we can consider a cover of $C$ by $X$-open sets or $Y$-open sets, it doesn't matter). If that is indeed your point, you're right, but it's not a real simplification IMHO. Munkres probably does this extra intersection because there might be other neighbourhoods of $\infty$ in $\mathcal{A}$ besides this one set $Y-C$ he has singled out.