Question about proof that an interval of rationals isn't a null set 
Let $X=\mathbb{Q}\cap[a,b]$, with $a<b$. This set isn't a null set
  because if it were, then given $\epsilon<b-a$, there would be a
  partition $P$ of $[a,b]$ such that the sum of the lenghts of the
  intervals of $P$ containing points of $X$ would be $< \epsilon$. Well,
  the sum of lenghts of all the intervals of $P$ is $b-a$, therefore
  some intervals of $P$ wouldn't contain points of $X$, that is,
  rational numbers, which is an absurd.

I didn't understand this proof. I can imagine open sets covering $[a,b]$ and all of them would have rationals contained in it, and their sum is $<\epsilon$. What am I not getting?
 A: There are two relevant notions of a "small" set of reals.
The first is measure:

A set $X\subseteq \mathbb{R}$ has measure zero (or is null) if $\forall\epsilon>0$, there is a family of open intervals $U_i$ ($i\in I$) such that $\bigcup_{i\in I} U_i\supseteq X$ and $\sum_{i\in I}\vert U_i\vert<\epsilon$ (where "$\vert U_i\vert$" denotes the length of $U_i$, not the cardinality).

Under this definition, the rationals and their subsets - and every countable set, more generally - is indeed null. For suppose $X=\{a_i: i\in\mathbb{N}\}$. Then for $\epsilon>0$, let $U_i=(a_i-2^{-i-17}, a_i+2^{-i-17})$.
However, note that this requires the family of open intervals to be infinite. If we restrict (as does the definition you mention in the comments) to finite families, we get a very different notion:

A set $X\subseteq\mathbb{R}$ is negligible (I think "has null content" is misleading here) if $\forall\epsilon>0$, there is a finite family of open intervals $U_i$ ($i\in I$) such that $\bigcup_{i\in I} U_i\supseteq X$ and $\sum_{i\in I}\vert U_i\vert<\epsilon$ (where again "$\vert U_i\vert$" denotes the length of $U_i$, not the cardinality).

Under this definition, countable sets need no longer be "small!" Indeed, $\mathbb{Q}$ and sets of the form $\mathbb{Q}\cap [a, b]$ for $a<b$ are not negligible. The proof is by induction: we show by induction on $n$ that any cover of $\mathbb{Q}\cap[a, b]$ by at most $n$ many open intervals has total length at least $b-a$, for every $n\in\mathbb{N}$.
That said, the proof you cite is terribly sloppy and unclear, and it skips over enough detail that I would say it is not correct.
