Lebesgue measure paradox See Lebesgue outer measure of $[0,1]\cap\mathbb{Q}$
Lebesgue measure:
$$ m(A) = \inf \left\{ \sum |I_n| : A \subset \bigcup I_n \right\} $$
We know that $m(\mathbb{Q} \cap [0,1]) = 0$.
Proof: 
Enumerate the rationals $\mathbb{Q} \cap [0,1] = \{ q_n \}_{n \in \mathbb{N}}$.
Let $A_n$ be the covering of rationals $\{ q_n \}_{n \in \mathbb{N}}$ in $[0,1]$ by intervals $A_n = (q_n - \varepsilon/2^n , q_n + \varepsilon/2^n)$. Then,
$$  m(\mathbb{Q} \cap [0,1]) \le m( \bigcup A_n ) \le \sum_n |A_n| = \sum_n \frac{\varepsilon}{2^n} = \varepsilon $$
and since it is true for every $\epsilon > 0$ as small as we want, we have  $m(\mathbb{Q} \cap [0,1]) = 0$. $\square$
Paradox?
Every number in $[0,1]$ is contained in an open neighborhood of a rational number. We reach a paradox if $$\bigcup_{n \in \mathbb{N}} A_n = [0,1] $$ 
since then 
$$m([0,1]) \le m( \bigcup A_n ) \le \sum m(A_n) = \varepsilon $$
and then $m([0,1])=0$ and not $1$ as it should be.
To contradict this we need to show
$$ \bigcup A_n \ne [0,1] \ . $$
We need to build an irrational number $r$ such that $\forall n : |r - q_n| > \varepsilon/2^n$. Can you construct such a number (or prove it exists without using Lebesgue measure argument)?
Moreover, I think we need to show the set $[0,1]-\bigcup A_n$ has $\aleph = 2^{\aleph_0}$ elements. Otherwise, $m([0,1]-\bigcup A_n) = 0$ and we reach a contradiction.
 A: If I may choose the enumeration of the rationals:
Enumerate the rationals in increasing order of the denominator in the reduced form, rationals with the same denominator in increasing order, so
$$q_1 = \frac01,\, q_2 = \frac11,\, q_3 = \frac12,\, q_4 = \frac13,\, q_5 = \frac23,\, q_6 = \frac14,\,q_7 = \frac34,\,\ldots$$
Then for $q_n = \frac{r_n}{s_n}$, we have $s_n \leqslant n$.
For $x = \frac{1}{\sqrt{2}}$, and a rational $0 \leqslant \frac{r}{s} \leqslant 1$, we have
$$\left\lvert\frac{r}{s} -\frac{1}{\sqrt{2}}\right\rvert = \frac{\left\lvert \frac{r^2}{s^2} - \frac12\right\rvert}{\left\lvert\frac{r}{s} +\frac{1}{\sqrt{2}}\right\rvert} \geqslant \frac{\lvert 2r^2-s^2\rvert}{4s^2} \geqslant \frac{1}{4s^2}.$$
For $0 < \varepsilon < \frac29$, we have $\frac{1}{4n^2} > \frac{\varepsilon}{2^n}$ for all $n \geqslant 1$, hence $1/\sqrt{2}$ is not contained in any $(q_n - \varepsilon 2^{-n},\,q_n + \varepsilon 2^{-n})$.
By a similar reasoning, we find $2^{\aleph_0}$ irrationals that aren't covered:
For an irrational $x \in (0,\,1)$ with continued fraction expansion
$$x = [a_0,\, a_1,\, a_2,\, \dotsc],$$
convergents $\frac{r_n}{s_n}$, and the complete quotients $\alpha_n$ - that means
$$x = \frac{\alpha_{n+1}r_n + r_{n-1}}{\alpha_{n+1}s_n + s_{n-1}},$$
and thus the continued fraction expansion $\alpha_n = [a_n,\, a_{n+1},\, \dotsc]$, whence $a_n < \alpha_n < a_n + 1$ - we have
$$\begin{align}
\left\lvert x - \frac{r_n}{s_n}\right\rvert &= \left\lvert \frac{\alpha_{n+1}r_n + r_{n-1}}{\alpha_{n+1}s_n + s_{n-1}} - \frac{r_n}{s_n}\right\rvert\\
&= \left\lvert \frac{\alpha_{n+1}(r_n s_n - s_nr_n) + (r_{n-1}s_n - r_ns_{n-1})}{(\alpha_{n+1}s_n+s_{n-1})s_n}\right\rvert\\
&= \left\lvert\frac{(-1)^{n}}{(\alpha_{n+1}s_n+s_{n-1})s_n}\right\rvert\\
&> \frac{1}{\bigl((a_{n+1}+1)s_n + s_{n-1}\bigr)s_n}\\
&\geqslant \frac{1}{(a_{n+1}+2)s_n^2}.\tag{1}
\end{align}$$
For a rational number $\frac{r}{s}$ with $s \leqslant s_n$, $n \geqslant 1$, we have
$$\lvert sx - r\rvert \geqslant \lvert s_nx - r_n\rvert$$
with equality if and only if $s = s_n$ and $r = r_n$.
This immediately leads to
$$\left\lvert x - \frac{r}{s}\right\rvert > \frac{1}{(a_{n+1}+2)s\cdot s_n}.\tag{2}$$
Thus, for any fixed $M \in \mathbb{Z}^+$, let $A_M$ be the set of all irrational numbers in $(0,\,1)$ whose continued fraction expansion has all partial quotients $\leqslant M$.
For $M \geqslant 2$, the set $A_M$ has cardinality $2^{\aleph_0}$.
For $x \in A_M$, the denominators of the convergents satisfy
$$s_{n+1} = a_{n+1}s_n + s_{n-1} \leqslant M\cdot s_n + s_{n-1} \leqslant (M+1)\cdot s_n.\tag{3}$$
Choosing the minimal $n$ with $s_n \geqslant s$ in $(2)$, we deduce
$$\left\lvert x - \frac{r}{s}\right\rvert > \frac{1}{(M+2)(M+1)s^2}$$
from $(3)$ for $x \in A_M$.
If $\varepsilon$ is small enough, so that
$$\frac{1}{(M+2)^2n^2} > \frac{\varepsilon}{2^n}$$
for all $n$, the set
$$U_\varepsilon = \bigcup_{n=1}^\infty \left(q_n - \frac{\varepsilon}{2^n},\,q_n + \frac{\varepsilon}{2^n}\right)$$
doesn't intersect $A_M$.
A: It will be difficult to prove the statement using topological arguments alone! Consider an analogous covering with wider intervals: $|B_n|=\epsilon/n$. Then $\bigcup B_n$ might or might not cover the whole interval $[0,1]$; it depends on the enumeration of rationals $\{q_n\}$.
The intervals $B_n$ look a lot like $A_n$. But in the situation $|A_n|=\epsilon/2^n$, we know that $\bigcup A_n$ doesn't cover the whole interval, regardless of the enumeration $\{q_n\}$, precisely because of a measure argument. I suspect that any proof of this fact will have to be a thinly disguised version of the same argument. (But by all means, prove me wrong!)
