What about using Lebesgue outer measure? The interval $[0,1]$ has Lebesgue outer measure 1, while countable sets have Lebesgue outer measure (the proof$0$.
For the purposes of this uses compactnessthe proof, I define the Lebesgue outer measure $\mathcal{L}(E)$ of a set $[0,1]$, which can be proved just from$E\subset\mathbb{R}$ as the completenessinfimum of the sums $\mathbb{R}$$\sum_i (b_i-a_i)$, where $E\subset \bigcup_i (a_i,b_i)$ (e.g. the infimum is over all countable coverings by open intervals).
On the other hand, itIt is a direct consequence of the definition that any countable set has Lebesgue outer measure 0. This can be even proved in the spirit of Gowers' first suggestion: suppose that $f:\mathbb{Q}\cap (0,1)\to A$ is a bijection. Then, given $\varepsilon>0$, the family $$\{ ( 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 $O(\varepsilon)$.
To prove that $\mathcal{L}([0,1])=1$, the following is the key claim: Let $\{ (a_i,b_i)\}$ be a finite cover of the interval $[c,d]$ with no proper subcover. Then $\sum_i (b_i-a_i) > d-c$.
The claim is proved by induction in the number of elements of the cover. It is clearly true if the cover has just one interval. Now if $[c,d] \subset \bigcup_{i=1}^n (a_i,b_i)$ with $n>1$, then $[c,d]\backslash (a_1,b_1)$ is either a closed interval $I$ or the union $I\cup I'$ of two disjoint closed intervals. In the first case $\bigcup_{i=2}^n (a_i,b_i)$ is a cover of $I$ and we apply the inductive hypothesis to it. Otherwise, $\{(a_i,b_i)\}_{i=2}^n$ can be split into two disjoint subfamilies, one which covers $I$ and one which covers $I'$. We then apply the inductive hypothesis to these families. (We use the property that the original cover has no proper subcover to make sure the covers of $I$ and $I'$ are disjoint.)
Now the claim and compactness of $[0,1]$ (ie. Heine-Borel) yield that $\mathcal{L}([0,1])\ge 1$.
Hence, $[0,1]$ is uncountable and so is $\mathbb{R}$.