The argument is basically just fine, but the notation in that last line is a bit off: when you write $\bigcup_{n\in\mathbb{N}}$, you’re saying that you want to take the union of some sets indexed by natural numbers. What follows $\bigcup_{n\in\mathbb{N}}$ should be a typical one of these sets with index $n$. If you knew that every member of $\mathcal{B}$ was compact, for instance, you could write $X = \bigcup_{n\in\mathbb{N}}\operatorname{cl}B_n$. What you want here is simply $X = \bigcup \{\operatorname{cl}B:B \in \mathcal{B}\text{ and}\operatorname{cl}B\text{ is compact}\}$.
If fact, you never needed to index $\mathcal{B}$ in the first place. It’s perfectly fine to say this:
- Let $\mathcal{B}$ be a countable base. Since $X$ is locally compact, for each $x\in X$ there is an open nbhd $U_x$ of $x$ with compact closure, and since $\mathcal{B}$ is a base for $X$, there is some $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq U_x$. Clearly each $B_x$ has compact closure, and $X = \bigcup\limits_{x\in X}\operatorname{cl}B_x$. There are only countably many distinct members of $\mathcal{B}$, so there are only countably many different sets $B_x$, and we have therefore expressed $X$ as the union of countably many compact subsets, as desired.
Or you could replace that last sentence with something like this:
- $\{B_x:x\in X\} \subseteq \{B \in \mathcal{B}:\operatorname{cl}B\text{ is compact}\} \subseteq \mathcal{B}$, so $\{B_x:x\in X\}$ is countable, and $X = \bigcup\limits_{x\in X}\operatorname{cl}B_x$ therefore expresses $X$ as a countable union of compact subsets.
There are many other perfectly good ways to do it; I’m just trying to give you some idea of what’s possible, since you’re working on the proof-writing as much as you are on the mathematics itself.
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.
Best Answer
Let $Y$ be an uncountable set, let $p$ be a point not in $Y$, and let $X=\{p\}\cup Y$. Let
$$\mathscr{B}=\{\{x\}:x\in Y\}\cup\{X\setminus F:F\text{ is a finite subset of }Y\}\;;$$
then $\mathscr{B}$ is a base for a Hausdorff, locally compact, compact topology $\tau$ on $X$ that is not second countable. It isn’t even first countable: $p$ has no countable local base. $\langle X,\tau\rangle$ is the one-point compactification of the discrete space $Y$. (Of course $Y$ itself is also an example, but it’s merely locally compact; $X$ is also compact.)
You can make examples with even nicer properties. For instance, $Y\times[0,1]$ is Hausdorff, locally compact, and locally connected but not second countable. The closed long ray has all of those properties and is in addition connected, path connected, and locally path connected. Its one-point compactification loses path connectedness but gains compactness.