A direct proof of the paracompactness of $\mathbb{Q}$ I know $\mathbb{Q}$ is metric, and thus paracompact. But is there a way to show that every covering has a locally finite refinement?
 A: Here is a direct proof that $\Bbb Q$ is paracompact.
Let $\mathscr{U}$ be an open cover of $\Bbb Q$; without loss of generality assume that the elements of $\mathscr{U}$ are open intervals. For $A\subseteq\Bbb Q$ let $\operatorname{st}(A,\mathscr{U})=\bigcup\{U\in\mathscr{U}:U\cap A\ne\varnothing\}$; write $\operatorname{st}(x,\mathscr{U})$ for $\operatorname{st}(\{x\},\mathscr{U})$. For $x\in\Bbb Q$ let $\operatorname{st}^1(x,\mathscr{U})=\operatorname{st}(x,\mathscr{U})$, and for $n\in\Bbb Z^+$ let $\operatorname{st}^{n+1}(x,\mathscr{U})=\operatorname{st}\big(\operatorname{st}^n(x,\mathscr{U}),\mathscr{U}\big)$. Finally, for $x\in\Bbb Q$ let $I(x)=\bigcup\{\operatorname{st}^n(x,\mathscr{U}):n\in\Bbb Z^+\}$. Clearly $I(x)$ is open and order-convex, and $y\in I(x)$ iff $x\in I(y)$, so $\mathscr{I}=\{I(x):x\in\Bbb Q\}$ is a partition of $\Bbb Q$ into clopen, order-convex sets.
Fix $x_0\in\Bbb Q$, let $L=(\leftarrow,x_0]\cap I(x_0)$, and let $R=[x_0,\to)\cap I(x_0)$. If $\operatorname{st}(x_0,\mathscr{U})\nsupseteq R$, there must be an $x_1\in R\cap\operatorname{st}^2(x_0,\mathscr{U})\setminus\operatorname{st}(x_0,\mathscr{U})$. Given $x_n\in R\cap\operatorname{st}^{n+1}(x_0,\mathscr{U})\setminus\operatorname{st}^n(x_0,\mathscr{U})$, if $\operatorname{st}^{n+1}(x_0,\mathscr{U})\nsupseteq R$, there is an $x_{n+1}\in R\cap\operatorname{st}^{n+2}(x_0,\mathscr{U})\setminus\operatorname{st}^{n+1}(x_0,\mathscr{U})$. There are now two cases.


*

*There is some $m_R\in\Bbb N$ such that $\operatorname{st}^{m_R+1}(x_0,\mathscr{U})\supseteq R$. Then for $n=0,\dots,m_R$ there are $U_n\in\mathscr{U}$ such that $x_n\in U_n$, and moreover $U_n\cap U_{n+1}\ne\varnothing$ if $n\ne m_R$; clearly $\{U_0,\dots,U_{m_R}\}$ covers $[x_0,x_{m_R}]$. If there is some $U_{m_R+1}\in\mathscr{U}$ such that $x_{m_R}\in U_{m_R+1}\supseteq R\cap[x_{m_R},\to)$, then $\{U_0,\dots,U_{m_R+1}\}$ is a finite and hence locally finite open cover of $R$ refining $\mathscr{U}$.  
Otherwise there must be a strictly increasing sequence $\langle z_n:n\in\Bbb N\rangle$ in $R\cap[x_{m_R},\to)$ and sets $V_n\in\mathscr{U}$ for $n\in\Bbb N$ such that $x_{m_R},z_n\in V_n$, $z_{n+1}\notin V_n$, and $R\cap[x_{m_R},\to)\subseteq\bigcup_{n\in\Bbb N}V_n$. Moreover, we may assume that $z_0>x_{m_R}$. Let $W_0=V_0\cap(x_{m_R},z_1)$, and for $n\in\Bbb Z^+$ let $W_n=V_n\cap(z_{n-1},z_{n+1})$. $\mathscr{W}=\{W_n:n\in\Bbb N\}$ refines $\mathscr{U}$ and covers $R\cap[x_{m_R},\to)$, and each member of $\mathscr{W}$ meets at most two other members of $\mathscr{W}$, so $\mathscr{W}\cup\{U_0,\dots,U_{m_R}\}$ is a locally finite open cover of $R$ refining $\mathscr{U}$.

*We end up with a sequence $\langle x_n:n\in\Bbb N\rangle$. Then for $n\in\Bbb N$ there are $U_n\in\mathscr{U}$ such that $x_n\in U_n$ and $U_n\cap U_{n+1}\ne\varnothing$. The choice of the points $x_n$ ensures that $U_m\cap U_n\ne\varnothing$ iff $|n-m|\le 1$, so $\{U_n:n\in\Bbb N\}$ is a locally finite open cover of $R$ refining $\mathscr{U}$.
Thus, in all cases we can find a locally finite open refinement $\mathscr{R}_R$ of $\mathscr{U}$ covering $R$. In similar fashion we can find a locally finite open refinement $\mathscr{R}_L$ of $\mathscr{U}$ covering $L$, and $\mathscr{R}=\mathscr{R}_L\cup\mathscr{R}_R$ is then a locally finite open refinement of $\mathscr{U}$ whose union is precisely $I(x_0)$.
Since we can do this for each $I\in\mathscr{I}$, $\mathscr{U}$ has a locally finite open refinement covering $\Bbb Q$, and $\Bbb Q$ is paracompact.
