Why must a locally compact second countable Hausdorff space be second countable to imply paracompactness? The textbook version of the result I've seen states: A locally compact second countable Hausdorff space is paracompact. Is the property of being second countable needed, or have I missed something? 
My thinking:


*

*If the space is locally compact then each point has a compact neighborhood.

*For this compact neighborhood each covering has a finite sub-covering.

*The finite sub-covering is a locally finite refinement. 
Thanks in advance.
 A: Overall the hypotheses are gross overkill, but you can’t omit any of them. Matt Samuel has already given the example of the long line to show that you can’t omit second countability. 
The particular point topology on a countably infinite set $X$ is non-Hausdorff, locally compact, and second countable — if $p$ is the particular point, $\mathscr{B}=\big\{\{p,x\}:x\in X\big\}$ is a countable base for the topology — and it’s not paracompact, since the open cover $\mathscr{B}$ has no open refinement that is locally finite at $p$. This shows that you can’t omit Hausdorffness.
To see that you can’t omit local compactness, let 
$$\begin{align*}
D&=\left\{\left\langle\frac1m,\frac1n\right\rangle:m,n\in\Bbb Z^+\right\}\;,\\\\
H&=\left\{\left\langle\frac1m,0\right\rangle:m\in\Bbb Z^+\right\}\;,\text{ and}\\\\
X&=\{\langle 0,0\rangle\}\cup H\cup D\;.
\end{align*}$$
Points of $D$ are isolated. For $m,n\in\Bbb Z^+$ let $x_m=\left\langle\frac1m,0\right\rangle$ and $y_{m,n}=\left\langle\frac1m,\frac1n\right\rangle$, and define
$$B_n(x_m)=\{x_m\}\cup\{y_{m,k}:k\ge n\}\;;$$
$\{B_n(x_m):n\in\Bbb Z^+\}$ is a local base at $x_m$. (In other words, the sequence $\langle y_{m,n}:n\in\Bbb Z^+\rangle$ converges to $x_m$.) Let $p=\langle 0,0\rangle$, and for $n\in\Bbb Z^+$ let 
$$B_m(p)=\{y_{k,n}:k\ge m\}\;;$$
$\{B_m(p):m\in\Bbb Z^+\}$ is a local base at $p$. With this topology $X$ is Hausdorff and second countable, but it’s not locally compact at $p$. It’s also not paracompact: the open cover 
$$\mathscr{U}=\{B_1(p)\}\cup\{B_1(x_n):n\in\Bbb Z^+\}$$
has no open refinement that is locally finite at $p$.
The only real function served by local compactness is to make the space regular: every locally compact Hausdorff space is regular. If we require the space to be regular and Hausdorff, we can drop the assumption of local compactness, because every second countable space is Lindelöf, and every $T_3$ Lindelöf space is paracompact. Second countability is of course a good deal stronger than the Lindelöf property, so this theorem is really just a weak version of the theorem that a $T_3$ Lindelöf space is paracompact.
A: Without the second countable condition the result is not true. The long line is locally compact Hausdorff and is not paracompact because it is locally metrizable but not metrizable.
