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What about using Lebesgue outer measure? The interval $[0,1]$ has Lebesgue outer measure 1(, while countable sets have Lebesgue outer measure $0$.

For the proof purposes of this uses compactness the 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}$as the completeness infimum of the sums$\mathbb{R}$). On \sum_i (b_i-a_i)$, where $E\subset \bigcup_i (a_i,b_i)$ (e.g. the other hand, it infimum is over all countable coverings by open intervals).

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 $$\{ ( f(p/q)-\varepsilon/q^3, f(p/q)+\varepsilon/q^3): p/q\in [0,1], \text{g.c.d.}(p,q)=1\}$$

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 $\mathcal{L}([0,1])\ge 1$.

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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. (Although this does use an enumeration This can be even proved in the spirit of a countable set so does not satisfy Gowers' first requirement.)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 $\mathbb{R}$.

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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. (Although this does use an enumeration of a countable set so does not satisfy Gowers' first requirement.)

Hence, $[0,1]$ is uncountable and so is $\mathbb{R}$.

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