You don’t even have to deal directly with rationals (except as radii of metric balls): you can prove the superficially more general result that a countable metric space without isolated points is not complete. (The generality is only superficial, because such a space is homeomorphic to $\Bbb Q$.)
Let $\langle X,d\rangle$ be a countable metric space without isolated points, and index $X=\{x_k:k\in\Bbb N\}$. Let $m_0=0$, $r_0=1$, $B_0=\{x\in X:d(x_{m_0},x)<r_0\}$, $C_0=\operatorname{cl}B_0$, and $M_0=\{k\in\Bbb N:x_k\in B_0\}$; $M_0$ is infinite, since $X$ has no isolated points, and $m_0=\min M_0$.
Suppose that $n\in\Bbb N$, and we have an $m_n\in\Bbb N$, an open set $B_n$ containing $x_{m_n}$, $C_n=\operatorname{cl}B_n$, and an infinite $M_n=\{k\in\Bbb N:x_k\in B_n\}$ such that $m_n=\min M_n$. Let
$$m_{n+1}=\min(M_n\setminus\{m_n\})\;;$$
then $x_{m_{n+1}}\in B_n\setminus\{x_{m_n}\}$. We can therefore choose a positive rational $$r_{n+1}\le\min\left\{d(x_{m_n},x_{m_{n+1}}),r_n-d(x_{m_n},x_{m_{n+1}}),2^{-(n+1)}\right\}$$ and set
$$\begin{align*}
&B_{n+1}=\{x\in X:d(x_{m_{n+1}},x)<r_{n+1}\}\;,\\
&C_{n+1}=\operatorname{cl}B_{n+1}\;,\text{ and}\\
&M_{n+1}=\{k\in\Bbb N:x_k\in B_{n+1}\}\;.
\end{align*}$$
It’s not hard to verify that $C_{n+1}\subseteq B_n$, $M_{n+1}$ is infinite, and $m_{n+1}=\min M_{n+1}$, so the recursive construction can continue. Note that we always have $m_{n+1}>m_n$.
In the end we have a sequence $\langle C_n:n\in\Bbb N\rangle$ of closed sets such that $C_n\supset C_{n+1}$ for each $n\in\Bbb N$. Moreover, $\operatorname{diam}C_n\le 2r_n\le2\cdot2^{-(n+1)}=2^{-n}$, so if $X$ were complete, the Baire category theorem would ensure that $\bigcap_{n\in\Bbb N}C_n=\varnothing$. Suppose that some $x_k\in\bigcap_{n\in\Bbb N}C_n$. Then $k\in\bigcap_{n\in\Bbb N}M_n$. But $\langle m_n:n\in\Bbb N\rangle$ is strictly increasing, so there is an $n\in\Bbb N$ such that $k<m_n=\min M_n$, and hence $x_k\notin B_n$. Finally, $C_{n+1}\subseteq B_n$, so $x_k\notin C_{n+1}$, and $X$ therefore cannot be complete.
