The set of irrational numbers in $[0,1]$ contains closed intervals only? Is the set of irrational numbers in $[0,1]$ contains closed intervals only? After all, the set of irrational numbers in any interval is uncountable and dense in the set of real numbers, so it has a measure of the interval itself. Can I develop a strict subset of set of the irrational numbers in $[0,1]$ with the difference of their measures less than an epsilon?
 A: You have mixed up several different ideas here. Firstly, the set of irrattionals in $[0,1]$ is not closed- in fact, the rationals are dense in $[0,1]$, so there are many missing limit points. However, as you note, the interval $I$ does have positive measure- in fact, given any set of zero measure $M$, in particular for any countable collection (like the rationals), the measure of $I-M$ is the same as the measure of $I$. So, if you take any countable collection, say the set of all rationals multiplied by some irrational number, and subtract that set from the irrationals between 0 and 1, the measure will still be the same.
Edit-
In response the the clarified question (from the comment above), there are both no zero measure closed intervals on the real line, and no closed intervals which are subsets of the irrationals, because the rationals are dense in R. Thus, not only can you not pick a collection of closed intervals in the irrationals, if you were to do so, their union would have strictly positive measure.
A: For the new question about intervals, the fact that the rationals are dense in $[0,1]$ says that every interval includes at least one (in fact countably infinitely many) rationals, so the irrationals do not include any interval.  You seem to be assuming that any set of full measure, which the irrationals are, must include the whole interval, but that is not correct and the irrationals are an excellent example.
A: Let $I=\{x\,:\, x\not\in\mathbb{Q}\cap[0,1]\}$ and let $\overline{I}$ be the closure of $I$. 
If $I$ were expressible as a union of closed intervals, then $I$ would be closed and $\overline{I}=I$. 
But we know that $I$ is dense in $[0,1]$ thus $\overline{I}=[0,1]\ne I$. 
