Doubt in proof of one point compactification of Locally Compact Hausdorff Space This theorem is Proved in Munkres' Topology which says:

Let $X$ be a space. Then $X$ is locally compact Hausdorff if and only if there exists a 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 $Y'$ are two spaces satisfying these conditions, then there is a homeomorphism of $Y$ with $Y'$ that equals the identity map on $X.$

He start constructing space $Y$ by defining topology on $Y$ for proving if part of theorem. Everything was smooth until he proves that the constructed set $Y$ is compact. He writes:


To show that $Y$ is compact, let $\mathcal{A}$ be an open covering of $Y$. The collection $\mathcal{A}$ must contain an open set of type (2)(All open set of form $Y-C$ where $C$ is compact subspace of $X$), say $Y − C$, since none of the open sets of type (1)(All open set of form U where U is open in $X$) contain the point $\infty$. Take all the members of $\mathcal{A}$ 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 $\mathcal{A}$ will, along with the element $Y − C$, cover all of $Y$.


The trouble is at last line where he says that finite collection of $\mathcal{A}$,along with $Y-C$ cover all of $Y$. I understood that finite elements of $\mathcal{A}$ cover $C$(Locally compact). Now, my question is which theorem guarantees that $Y-C$ has finite subcovering? $C$ is locally compact so given an open covering of $Y$ which $Y-C$ will have only finite subcovering? This may be easy but I think I have missed something here. Thanks for the help.
 A: You need to re-read Munkres. It is a long proof, and it is (probably) hard to get it in one go.
I will (more-or-less) copy the proof here from Munkres. We want to show that $Y$ is compact. So we start with an open cover of $Y,$ say $\mathcal{A}.$
Since $\mathcal{A}$ covers $Y,$ every point of $Y$ must be contained in some open set in the family $\mathcal{A}.$ In particular, there is some $U\in \mathcal{A}$ which contains infinity. At this point we need to be a bit careful, and think about open sets of $Y$ which contain $\infty.$ To make matter smooth, Munkres has already grouped the open sets of $Y$ into two 'types'. The first type of open sets are contained entirely inside $X$ and hence can not contain $\infty.$ The second type of open sets have a `beautiful look'. They look like $Y\setminus C$ for some compact set $C\subseteq X.$ What this tells us is that, our friend $U$ must look like $Y\setminus C$ for some compact set $C$ (and recall that $U\in \mathcal{A}$).
The upshot is that in any cover of $Y$ there must be at least one 'very large' open set which covers almost whole $Y$ (in the sense, it leaves behind only a compact set which let us agree to call small for the time being).
Now rest of the set in $\mathcal{A}$ actually form a cover of $C$ which is compact. So you might want to say that I can choose only finitely many of them to do the job. Well, you (almost) can. The only caveat is that you need to cover $C$ by open subsets of $X$ not by open subsets of $Y.$ To get around this, you just need to note that any open subset of $Y$ can be intersected with $X$ to get an open subset of $X$ (why?). Therefore, the honest argument would be to say that, we already have one distinguished open set $U=Y\setminus C\in \mathcal{A}$ and if you intersect every remaining open set in this cover with $X$ it gives a cover of $C$ and hence you can find a finite subcover which still covers $C.$ These open sets (actually their original father, that is open set of $Y$ from which it was obtained after intersecting with $X$) together with our distinguished $U$ forms a finite cover of whole space $Y.$
I think the best way to understand it is to draw some pictures and visualize how open sets look like in $Y,$ and then try proving this result on your own. 
A: Munkres has shown that the collection of sets of type (1) and of type (2) forms an topology on $Y$. Given an open cover $\mathcal A$ of $Y$, one of the elements of $\mathcal A$ must contain $\infty$ and it must have the form $A_\infty = Y \setminus C$ with a compact $C \subset X$. Now $\mathcal A$ is a cover of $C$ by open sets in $Y$, thus there exist finitely many $A_1,\ldots,A_n \in \mathcal A$ such that $C \subset \bigcup_i A_i$ (see Lemma 26.1 in Munkres). But then $A_1,\ldots,A_n, A_\infty$ is a finite cover of $Y$.
