Theorem 29.1 of Munkres’ Topology 
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 $X$.

My attempt:
We first verify uniqueness. Claim: $h:Y\to Y’$ defined by $h(x)=x$, $\forall x\in X$ and $h(p)=q$, where $p\in Y-X$ and $q\in Y’-X$, is homeomorphism. Proof: It’s easy to show $h$ is bijective. Let $U\in \mathcal{T}_Y=\mathcal{T}_X \cup \{Y-C| C\subseteq X$ and $C$ is compact$\}$. We show $h(U)\in \mathcal{T}_{Y’}$. If $p\notin U$, then $U\in \mathcal{T}_X$. Assume towards contradiction, $U=Y-D$; $D\subseteq X$ and $D$ is compact. Then $p \in Y-D=U$. Which contradicts our initial assumption. Thus $U\in \mathcal{T}_X \subseteq \wp({X})$. So $U\subseteq X$. By definition of $h$, $h(U)=U$. Hence $h(U)=U\in \mathcal{T}_X \subseteq \mathcal{T}_X \cup \{Y’ -C| C\subseteq X$ and $C$ is compact$\} = \mathcal{T}_{Y’}$. Now If $p\in U$, then $U\in \{Y -C| C\subseteq X$ and $C$ is compact$\}$. Assume towards contradiction, $U\in \mathcal{T}_X$. Then $U\subseteq X$. So $p\notin U$. Which contradicts our initial assumption. Thus $U= Y-D$; $D$ is compact subspace of $X$. It is easy to check $h(U)=h(Y-D)=Y’-D$ by elementary set theory. Hence $h(U)=Y’-D \in \mathcal{T}_{Y’}= \mathcal{T}_X \cup \{Y’ -C| C\subseteq X$ and $C$ is compact$\} $. This completes the proof of $h^{-1}=g$ is continuous. Proof of $h$ is continuous is very similar. Thus map $h$ is homeomorphism. Is this proof better version of Munkres proof?
Can you explain: (1)$U_1 \cap (Y-C_1)=U_1 \cap (X-C_1)$. (2) $U\cup (Y-C)=Y-(C-U)$ which is of type (2) because $C-U$ is a closed subspace of $C$ and therefore compact.
Honestly I don’t really understand the proof of $Y$ is compact. Let $\mathscr{A}=\{ A_\alpha |\alpha \in J\}$ be an open cover of $Y$. Then $\exists Y-C\in \mathcal{T}_Y$ such that $Y-C \in\mathscr{A}$. $C\cup (Y-C)=Y= \bigcup_{\alpha \in J} A_\alpha$. So $C\subseteq \bigcup_{\alpha \in J} A_\alpha$ $\Rightarrow$ $C\cap X=C \subseteq (\bigcup_{\alpha \in J} A_\alpha)\cap X= \bigcup_{\alpha \in J} (A_\alpha \cap X)$. So $\{ X\cap A_\alpha \in \mathcal{T}_X| \alpha \in J\}$ is an open cover $C$. Since $C$ is compact, $\exists \{X\cap A_{\alpha_1},…, X\cap A_{\alpha_n}\}$ finite subcover. How to show $C\subseteq \bigcup_{k=1}^{n} A_{\alpha_k}$?
If $X$ is compact in place of locally compact, then space $Y=X\cup \{\infty\}$ equipped with $\mathcal{T}_Y=\mathcal{T}_X \cup \{\infty\}$ topology satisfy all three condition. Am I right?
Finally, If $X$ is not compact, then $\overline{X}=Y$. Proof: we need to show $\infty \in \overline{X}$. Let $V\in \mathcal{N}_\infty$. So $\infty \in V$ and $V\in \mathcal{T}_Y$. If $V\in \mathcal{T}_X$, then $\infty \notin V$. Which contradicts Our initial assumption. So $V\in \{ Y-C|C\subseteq X$ and $C$ is compact $\}$. $V=Y-D$; $D\subseteq X$ and $D$ is compact. By hypothesis, $D\neq X$. So $D\subset X$. $\exists x\in X-D$. $X\subset Y \Rightarrow X-D\subset Y-D$. So $x\in Y-D=V$ and $x\in X$. Hence $x\in V\cap X\neq \emptyset$. $\infty \in \overline{X}$. Is this proof correct?
 A: *

*Unfortunately your uniqueness proof is not better than Munkres'. Actually it is longer and a bit confusing. When you say "Let $U\in \mathcal{T}_Y=\mathcal{T}_X \cup \{Y-C| C\subseteq X$ and $C$ is compact$\}$", it seems that you claim that it is already known that the topology on $Y$ has the form  $\mathcal{T}_Y=\mathcal{T}_X \cup \{Y-C| C\subseteq X$  and $C$ is compact$\}$. But this has to be proved. Munkres does it by observing that open subsets of $Y$ which do not contain $p$ are open subsets of $X$ (i.e. belong to $\mathcal{T}_X$) and open subsets of $Y$ which contain $p$ have the form $Y-C$ with compact $C \subseteq X$. Next you say "If $p\notin U$, then $U\in \mathcal{T}_X$. Assume towards contradiction, $U=Y-D$; $D\subseteq X$ and $D$ is compact. Then $p \in Y-D=U$. Which contradicts our initial assumption. Thus $U\in \mathcal{T}_X \subseteq \wp({X})$. So $U\subseteq X$." But this does not require a proof, if $p \notin U$, then trivially $U \in \mathcal{T}_X$. A similar remark concerns the case  $p\in U$.


*$U_1 \cap (Y-C_1)=U_1 \cap (X-C_1)$: This is obvious because $U_1 \in \mathcal T_X$ does not contain $p$ and $Y = X \cup \{p\}$, i.e. $Y-C_1 = (X-C_1) \cup \{p\}$.


*$U\cup (Y-C)=Y-(C-U)$ is of type (2):  For $y \in Y$  we have $y  \in U\cup (Y-C)$ iff $y \in U$ or $y \notin C$, and $y \in Y-(C-U)$ iff $y\notin C - U$ which means $y \notin C$ or $y \in U$. This proves the equation. I think it should be clear that $C-U = C \cap (X - U)$ is closed in $C$ and therefore compact.


*Since $\{X\cap A_{\alpha_1},…, X\cap A_{\alpha_n}\}$ is a cover of $C$ and trivially $X \cap A_{\alpha_k} \subseteq A_{\alpha_k}$ , we have $C = \bigcup (X \cap A_{\alpha_k}) \subseteq \bigcup A_{\alpha_k}$.


*$\mathcal{T}_X \cup \{\infty\}$ is not a topology unless $X = \emptyset$. Certainly $\{\infty\}$ is open in $Y$, and this means that all $U \cup \{\infty\}$ with $U \in \mathcal{T}_X$ are open in $Y$. Thus the correct version is $\mathcal{T}_Y=\mathcal{T}_X \cup \{U \cup \{\infty\} \mid U \in \mathcal{T}_X \}$.


*Your proof that $ \bar X = Y$ is correct.
