What about using Lebesgue outer measure? The interval $[0,1]$ has Lebesgue outer measure 1 (the proof of this uses compactness of $[0,1]$, which can be proved just from the completeness of $\mathbb{R}$).
On the other hand, it is a direct consequence of the definition that any countable set has Lebesgue outer measure 0. This can be even proved in the spirit of Gowers' first suggestion: suppose that $f:\mathbb{Q}\cap (0,1)\to A$ is a bijection. Then, given $\varepsilon>0$, the family $$\{ ( f(p/q)-\varepsilon/q^3, f(p/q)+\varepsilon/q^3): p/q\in [0,1], \text{g.c.d.}(p,q)=1\}$$ is a cover of $A$ by intervals, such that the sum of the lengths is $O(\varepsilon)$.
Hence, $[0,1]$ is uncountable and so is $\mathbb{R}$.