Does there exist a metric $d$ on $\mathbb{Q}$ such that $(\mathbb{Q},d)$ is compact?
EDIT
NOT considering $\mathbb{Q}$ as a subspace of $\mathbb{R}$. Can there exist a metric on $\mathbb{Q}$ which makes it compact? We may decide later what does it do to $\mathbb{R}$.
Here is a slightly more explicit take on Tsemo's answer:
Define the injective map $\phi:\mathbb{Q} \to \mathbb{C}$ by $\phi(0) = 0$ and if $q = { a \over b}$ where $a,b$ are coprime and $a \neq 0, b>0$ then $\phi({a \over b}) = {1 \over b} (-1 + e^{i {7 \over 8} \pi {1 \over a}})$. Let $d(q_1,q_2) = |\phi(q_1)-\phi(q_2)|$.
The following crude picture might help illustrate:
Two important characteristics of $\phi$ are that $\lim_{|a| \to \infty} \phi({a \over b}) = 0$ and $\lim_{b \to \infty} \phi({a \over b}) = 0$.
It is straightforward to verify that $d$ is indeed a metric.
It is sufficient to show that $(\mathbb{Q},d)$ is sequentially compact.
Suppose $q_n \in \mathbb{Q}$. Let $q_n = {a_n \over b_n}$ with $b_n \ge 1$, and $a_n,b_n$ coprime.
If $a_n = 0$ infinitely often then it is clear that there is a convergent subsequence (to zero), so we can suppose $|a_n| \ge 1$.
If $(a_n,b_n)$ is bounded in the usual metric then it is clear that $(a_n,b_n)$ has some subsequence that converges to some $(a,b)$ (in fact we have $(a_n,b_n) = (a,b)$ infinitely often).
Otherwise either $|a_n|$ or $b_n$ is unbounded, and hence any corresponding subsequence converges to zero.
Note in particular that if the $q_n$ are all distinct then $q_n \to 0$.