Closed and open subsets of locally compact Hausdorff space are locally compact These questions have been asked to death, but I found the proof for "open subsets of locally compact Hausdorff space are locally compact" too tedious in each of the answers I have sampled on MSE, for example:
In a locally compact Hausdorff space, why are open subsets locally compact?
Open subspaces of locally compact Hausdorff spaces are locally compact
(and nobody accept answers. Why??)
I think the proof for "closed subsets of locally compact Hausdorff space are locally compact" is very brief and elegant. I want to obtain a similar proof for open subsets.
The definition of locally compact I am sing is:
A space $(X, \mathfrak{T})$ is locally compact if $\forall x \in X$, $\exists K, U \subseteq X, K$ is compact, $U$ is open, s.t. $x \in U \subseteq K$
(Proof 1: Closed subsets of locally compact Hausdorff space are locally compact)
Proof:


*

*Let $(X,\mathfrak{T})$ be a locally compact Hausdorff space. 

*Then $\forall x \in X, \exists K$, $K$ compact, $U \in \mathfrak{T}$ s.t. $ x \in U \subseteq K$. 

*Let $C$ be closed, then $\forall x \in C, x \in U \cap C \subseteq C \cap K \subseteq C$. 

*Since $U \cap C$ is open, $C \cap K$ is compact, therefore $C$ is locally Hausdorff. The end. 

However, I can't figure out why I cannot complete the proof for "open subsets ..." in a few lines similar to above. 
(Proof 2: Open subsets of locally compact Hausdorff space are locally compact)
Proof Attempt:


*

*Let $(X,\mathfrak{T})$ be a locally compact Hausdorff space. 

*Then $\forall x \in X, \exists K$, $K$ compact, $U \in \mathfrak{T}$ s.t. $ x \in U \subseteq K$.  

*Let $V$ be closed, then $\forall x \in V, x \in U \cap V$
Now we just need to produce a compact set containing $U\cap V$ that is contained in $V$
I think to proceed I need to use the following Lemma:
Lem:

Given $(X, \mathfrak{T})$ Hausdorff, then it is locally compact iff
  every point is contained in an open set with compact closure

Then $\forall x \in V, x \in U \cap V \subset \overline{ (U \cap V)} \subseteq V$

All I am left to do is to show that $\overline{ (U \cap V)} \subseteq V$. Can anyone provide suggestion as to how to show this?
 A: Suppose $X$ is a locally compact Hausdorff space and $Y \in \Gamma(X)$. I'll prove that $Y$ is a locally compact Hausdorff space with the subspace topology.
I
Suppose $x,y \in Y$ and $x \not= y$.

*

*Then $\exists V_i \in \Gamma(X),i \in V_i$ where $i=x,y$ and $V_x \cap V_y = \phi$. This follows from X being Hausdorff.

*For $i=x,y$, put $G_i = Y \cap V_i$. Then $G_i \in \Gamma(Y)$, $i \in G_i$ and $G_x \cap G_y=\phi$.

Hence $Y$ is Hausdorff.
II
Let $x \in Y$. We have:

*

*$\{x\} \subset X$. $\{x\}$ is $X-$compact.

*By theorem 2.7 of RCA Rudin:
$\;\;\;\;\exists V \in \Gamma(X), \{x\} \subset V \
\subset cl_X(V) \subset Y \
\text{ such that } cl_X(V)\text{ is $X-$compact.}$


*$cl_X(V)$ is $Y-$compact as well. Also $V \in \Gamma(Y)$


*As $cl_X(V) = cl_X(V) \cap Y$, $cl_X(V)$ is $Y-$closed.


*By 2 and 4, $cl_Y(V) \subset cl_X(V)$.


*As closed subsets of compact sets are compact,$cl_Y(V)$ is $Y-$compact.
Now we have all the ingredients ready. We have:

*

*$V \in \Gamma(Y)$ such that $x \in V$.

*$cl_Y(V)$ is $Y-$compact.
Therefore Y is locally compact.

A: Ok, so we have a locally compact Hausdorff space $X$, and we want to show that any $U \in \tau_{_X}$ is also locally compact in the subspace topology.  By the lemma you cite, it suffices to show that for each $x \in U$, there is a compact neighborhood of $x$ that is contained entirely inside $U$.  
Since $x$ is already guaranteed a compact neighborhood $K \subset X$ due to local compactness, this will serve as our starting point. The intersection of a closed set and a compact set is itself compact, so our strategy going forward will be to find a (finite) collection of closed neighborhoods $\{ C_i \}$ of $x$ together with $K$ such that their intersection is a (necessarily compact) proper subset of $U$.
A good step in the right direction is to take $K \cap \overline{U}$, which clips off a ton of "extra" points in $K \setminus U$.  
To get a third closed set $S$ so that $S \cap K \cap \overline{U} \subset U$, we can do the following:  note that $\partial U \cap K$ is compact because $\partial U$ is closed (where $\partial U$ denotes the boundary of $U$), and since $X$ is Hausdorff, we can cover $\partial U\cap K$ with open sets that are "far" from $x$.  Let $T$ be the union of the open sets contained in the finite subcover this will necessarily admit, and let $S = T^c$ (which is a closed neighborhood of $x$).
$S \cap K \cap \overline{U}$ is a compact neighborhood of $x$ contained in $U$.

Footnote:  Of course, we don't need to be laborious above w.r.t. compactness when transitioning between the normal topology and the subspace topology on $U$ since if $S$ compact in a topological space $X$, then $S \cap Y$ is compact in any given subset $Y$ under the subspace topology.
