Every locally compact, second countable Hausdorff space has a countable basis of open sets with compact closure Let $X$ be a locally compact, second countable Hausdorff space. I want to prove that it has a countable basis of opens with compact closure, and that this basis can be extracted as a subset of any basis of its topology by restricting to opens with compact closure.
I understand the following:
As locally compact space, for every point $x$ of $X$ we can find a compact neighbourhood $V$, thus, for every point $x$ of $X$ we can find an open $V_x$ containing $x$ and whose closure is compact (by taking $V_x = V$ if $V$ is open, or the open set whose closure is $V$ if $V$ is closed, i.e: $V_x$ such that $V_x = V°$).
As second countable space, there is a basis family $\{U_{\alpha}\}_{\alpha \in A}$ of opens such that for every point $x$ of $X$, and every open $V_x$ containing $x$, there is a $\alpha\in A$ such that $x\in U_{\alpha}\subset V_x$.
Now, take a point $x$ in $X$, we can find an open $V_x$ around $x$ whose closure is compact, and find an element $U_{\alpha}$ of the basis such that $x\in U_{\alpha}\subset V_x$, the closure of $U_{\alpha}$ is a closed subset of the compact closure of $V_x$, thus, it is also compact.
Thus, for each point $x$ and for each open $V_x$ around that point whose closure is compact we can find an open $U$ with compact closure containing $x$ and contained in $V$.

Opens of the family basis whose closures are compact form a new basis family for the opens of the topological space with compact closures.

Now, how to show that it actually form a new basis family for all the opens of the topological space? I guess that the "Hausdorff" condition has something to do with.
 A: 
I want to prove that it has a countable basis of opens with compact closure,

That part is easy,

and that this basis can be extracted as a subset of any basis of its topology by restricting to opens with compact closure.

and this part is wrong as written, given the countable basis of the topology of $\mathbb{R}$ formed by the intervals with rational endpoints, we can easily construct an uncountable family of pairwise disjoint countable bases of the topology by shifting all intervals by a fixed irrational amount.
For the easy part, we formulate the
Lemma: Let $(X,\,\tau)$ be a locally compact Hausdorff space. For any basis $\mathcal{B}$ of $\tau$, the family $\mathcal{B}_c = \{ U \in \mathcal{B} : \overline{U} \text{ is compact}\}$ is also a basis of $\tau$.
Proof: Since $\mathcal{B}_c \subset \tau$, we only need to show
$$V \in \tau \Rightarrow V = \bigcup_{U \in \mathcal{B}_c,\, U \subset V} U,$$
or, that for every open $V$ and every $x \in V$, there is a $U \in \mathcal{B}_c$ with $x \in U \subset V$. Now, $\tau$ is locally compact, hence there is a compact neighbourhood $K_x$ of $x$ contained in $V$(1). Since $\mathcal{B}$ is a basis of $\tau$, there is a $U_x \in \mathcal{B}$ with $x \in U_x \subset \overset{\circ}{K}_x$. Since $\tau$ is Hausdorff, compact sets are closed, hence $\overline{U}_x \subset K_x$ is also compact, and therefore $U_x \in \mathcal{B}_c$. $\hphantom{foobarbazquuxbongbambooooopppppppppppppp}$ c.q.f.d.
If we start with a countable basis of the topology, extracting the relatively compact basis sets yields a countable basis of relatively compact sets.
Now the question remains whether in a second countable locally compact Hausdorff space every basis of the topology contains a countable basis.
That is the case for every second countable space (thanks to Stefan H. for the simple general proof).
Let $\mathcal{C}$ be a countable basis, and $\mathcal{B}$ any basis of the topology. Consider the set
$$\Gamma = \{ (C_1,\, C_2) \in \mathcal{C}^2 : (\exists B \in \mathcal{B})(C_1 \subset B \subset C_2)\}.$$
As a subset of the countable set $\mathcal{C}^2$, it is countable. For every $p = (C_1,\,C_2) \in \Gamma$, choose a $B_p \in \mathcal{B}$ with $C_1 \subset B_p \subset C_2$. Then $\mathcal{B}_\Gamma = \{ B_p : p \in \Gamma\}$ is a countable subset of $\mathcal{B}$. It is also a basis of the topology, since for open $V \neq \varnothing$ and $x \in V$, we can find a $C_2 \in \mathcal{C}$ with $x \in C_2 \subset V$, since $\mathcal{C}$ is a basis. Since $\mathcal{B}$ is a basis, there is a $B \in \mathcal{B}$ with $x\in B \subset C_2$. Since $\mathcal{C}$ is a basis, there is a $C_1 \in \mathcal{C}$ with $x \in C_1 \subset B$. Then $(C_1,\,C_2) \in \Gamma$, and hence
$$x \in C_1 \subset B_{(C_1,\,C_2)} \subset C_2 \subset V.$$
Thus, in a second countable locally compact Hausdorff space, every basis of the topology contains a countable basis of relatively compact sets.

(1) In a locally compact Hausdorff space $X$, every point has a neighbourhood basis of compact sets.
For $x \in X$, let $K_x$ be a compact neighbourhood of $x$.
Let $V$ be an arbitrary open neighbourhood of $x$. Then $U = V \cap \overset{\circ}{K}_x$ is an open neighbourhood whose closure is compact (since it is contained in $K_x$, which is closed because $X$ is Hausdorff). Thus $\partial U$ is a compact set, and $x \notin \partial U$. For every $y \in \partial U$, there are disjoint open neighbourhoods $W_y$ of $x$ and $Z_y$ of $y$. $\partial U$ is compact, hence there are $y_1,\, \ldots,\,y_k$ such that $\partial U \subset Z := \bigcup_{j=1}^k Z_{y_j}$. $W := \bigcap_{j=1}^k W_{y_j} \cap U$ is an open neighbourhood of $x$ with $W \cap Z = \varnothing$, and $W \subset U$. Hence $\overline{W} \subset \overline{U}\setminus Z \subset \overline{U}\setminus \partial U = U \subset V$.
A: I don't usually think about such questions, but this appears to me to be a simplification of Daniel Fischer's proof of the first fact:
Let $\mathcal{B}$ be a countable basis. It suffices to show that for each open set $V$ and each $x \in V$ there is a relatively compact $W$ in $\mathcal{B}$ such that $x \in W \subset V$.
Now given such an open $V$ and $x \in V$, local compactness ensures that there is a compact $K$ such that $x \in K$ and $x \in K^{o}$. Consider $ U = V \cap K^{o}$. Since $U$  is open and $\mathcal{B}$ is a basis there exists some $W \in \mathcal{B}$ such that $x \in W \subset U$. Since $W \subset K^{o}$ we have that $\bar{W} \subset \bar{K^{o}}$, and since the space is Hausdorff and $K$ is compact $\bar{K^{o}} = K$. Thus $\bar{W}$ is   a closed subset of a compact set, and therefore compact. Thus $W$ is relatively compact, completing the proof.
