What is a metric for $\mathbb Q$ in the lower limit topology?

Let $\nu:\mathbb{Q} \to \mathbb{N}$ be an enumeration of $\mathbb{Q}$. Then $\displaystyle d(x,y):=\sum_{\min(x,y) < r \le \max(x,y)} 2^{-\nu(r)}$ will do.

EDIT: This one should work out.

Here's another option: write your rationals as "mixed fractions," that is as their integer floor plus a fractional part and define $$d\left(a\frac{p}{q},b\frac{r}{s}\right)=\begin{cases}|a-b|,& a\neq b\\ d'\left(\frac{p}{q},\frac{r}{s}\right), &\textrm{otherwise} \end{cases}.$$

Define the distance between pure fractions separately. Assume WLOG that $\frac{p}{q}\leq \frac{r}{s}$. Then set $$d'\left(\frac{p}{q},\frac{r}{s}\right)=\max\left(\left|\frac{p}{q}-\frac{r}{s}\right|,\frac{1}{m}\right), \frac{k}{m} \in \left(\frac{p}{q},\frac{r}{s}\right]$$ We see, for instance, that this gives $\left[\frac{1}{2},\frac{5}{6}\right)$ open, since the distance from anything smaller than $\frac{1}{2}$ to $\frac{1}{2}$ is $\frac{1}{2},$ but each distance from $\frac{1}{2}$ to something less than $\frac{5}{6}$ is no more than $\frac{1}{3}$.

The symmetry and homogeneity axioms are immediate. Let's consider the triangle inequality. The only case in which it doesn't follow from that for the standard absolute metric is when we must show $d'\left(\frac{p}{q},\frac{r}{s}\right)+d'\left(\frac{r}{s},\frac{t}{u}\right)\geq d'\left(\frac{p}{q},\frac{t}{u}\right)$ and $d'\left(\frac{p}{q},\frac{t}{u}\right)=\frac{1}{m}$. But if the inequality failed, we'd have to have the interval $\left(\frac{p}{q},\frac{t}{u}\right]$ contain something with denominator $m$ bigger than any denominator in the intervals on the left-hand side-which is absurd, since depending on the ordering we have one of $ \left(\frac{p}{q},\frac{t}{u}\right]=\left(\frac{p}{q},\frac{r}{s}\right]\cup\left(\frac{r}{s},\frac{t}{u}\right], \left(\frac{p}{q},\frac{t}{u}\right]\subset \left(\frac{p}{q},\frac{r}{s}\right],$ or $\left(\frac{p}{q},\frac{t}{u}\right]\subset\left(\frac{r}{s},\frac{t}{u}\right]$.

So we have a metric. To get some half-open interval $[x,y)=\left[a\frac{p}{q},b\frac{r}{s}\right)$, take the union of all the $[m,m+1) \subset [a,b)$. Then for the least such $m$, construct $[x,m)$ by the infinite union $\bigcup_{i=0}^\infty\left[a\frac{p}{q+i},a\frac{p}{q+i}+\frac{1}{q+i+1}\right)$, getting all these intervals by the argument above about $\left[\frac{1}{2},\frac{2}{3}\right)$. Get $\left[b,b\frac{r}{s}\right)$ as the ball of radius $\frac{r}{s}$ around $b$, union the three pieces together, and we're done.