Closed, uncountable set without rationals. Problem: Find a closed, uncountable without rationals,  subset in the unit interval. 
Solution: Let X be any binary irrational number , replace the 1s with 2s in the decimal expansion . 
Now we can use the digits of X as an index for a path through the cantor set, for example, if the first two digits of X are 0 2 then we take [0 , 1/3] and then we take [2/9 , 1/3] . 
This subset is obviously closed and uncountable, and since we are using an irrational index all the elements of that subset will be irrational. 
Ok, is there a big mistake here? And secondly, If all the end points of a cantor set are rational doesn't this contradict the above?
 A: I think what you want is the set  of those $t \in [0,1]$ with base $3$ expansion $.t_1 t_2 t_3 \ldots $ (i.e. $t = \sum_{j=1}^\infty t_j 3^{-j}$) such that
$t_j \in \{0,2\}$ if $x_j = 0$ and $t_j = 1$ if $x_j = 1$, where $X = .x_1 x_2 x_3 \ldots$ is the base $2$ expansion of the irrational number $X$. That would be a correct answer.
A: Not that you asked about it, but you can avoid the Cantor set-ternary expansion bit by considering any $U\subset [0,1],$ $U$ open in $[0,1],$ containing $\mathbb Q \cap [0,1],$ with $m(U)<1/2.$ Then $[0,1]\setminus U$ is a closed subset of $[0,1]$ having measure greater than $1/2$ that contains no rational.
If you know about the regularity of Lebesgue measure, then let $E$ be the set of irrational numbers in $[0,1].$ We have $m(E)=1$ and by regularity,
$$m(E) = \sup \{ m(K): K \subset E, K \text { closed in } \mathbb R \}.$$
It follows that there are plenty of of closed subsets of $[0,1]$ having positive measure that contain only irrational numbers.
A: 
This subset is obviously closed and uncountable, and since we are using an irrational index all the elements of that subset will be irrational.

The word "obviously" is one you should always be afraid of in math. While it is indeed obvious that your set is closed (being an intersection of closed sets) and it's not hard to show that it consists only of irrationals (although that isn't really trivial - you should say a bit about why it's true), it is not true that the set is uncountable!
I explained this in the case when $X=000...$ in the comments; your response was 

that can't happen since $X$ is irrational.

However, it makes no difference what $X$ is; no matter what we pick for $X$, we'll always get a single point. In fact, it's easy to describe the point you'll get: it's just $X$ itself, but with each $1$ replaced by a $2$ and then interpreted as a ternary expansion. E.g. if we have $$X=0.010011000111...$$ your process will produce the set $$\{0.020022000222...\}.$$
(Don't believe me? What's a different point in the set you get from this $X$?)
To see why this is true, just note that at each step of the process you're determining another digit of the elements in the set you're building. E.g. with the $X$ above, by looking at the first three digits it's clear that in the set you build, every point's ternary expansion will begin "$0.020...$". So your whole process determines the point completely. The reason the Cantor set is uncountable is that you "keep having choices" - e.g. your first ternary digit could be $0$ or $2$, your second ternary digit could be $0$ or $2$, etc. By removing these choices as you've done, you get only one point.
A: The Cantor set $C \subset [0,1]$ is an uncountable disjoint union of sets $C_x, x \in C$ that are all homomorphic to $C$ (so perfect compact etc.). There can only be a rational in countably many of the $C_x$ so lots of them contain not rationals.
The union property is direct from $C \times C \simeq C$ which can be seen from the ternary expansion representation of the Cantor set: split the expansion of $x \in C$ into an odd and even indexed part to produce a pair of points in $C$. Merging two such expansion is the inverse. 
An alternative but less elementary way to see the homeomorphism is to use Brouwer's theorem that all compact totally disconnected metric spaces without isolated points are homeomorphic to (each other and to )$C$.
A: Here is the simplest solution to your question.
$
\def\nn{\mathbb{N}}
\def\less{\smallsetminus}
$
Let $(r_n)_{n\in\nn}$ be a sequence enumerating all the rationals in $[0,1]$.
Let $S = [0,1] \less \bigcup_{n\in\nn} B(r_n,2^{-3-n})$, where $B(r,d)$ is the open interval centred at $r$ with radius $d$.
Then $S$ is clearly closed and has measure at least $1-\sum_{n=0}^\infty 2^{-2-n} = \frac12$.
A: I'm not sure if this is explicit enough for you:
Fix $0< \epsilon < 1$. Enumerate the rationals in $[0,1]$ by $\{q_n\}_{n \in \mathbb{N}}$. Around $q_n$ take the open ball $B(q_n, \epsilon 2^{-n})$.
Then $$[0,1] \setminus \bigcup_{n \in \mathbb{N}} B(q_n, \epsilon 2^{-n})$$ satsifies the properties you want. 
It is evidently closed, and it must be uncountable, since we are removing (at most) a collection of intervals of total length $$\sum_{n=1}^\infty 2^{-n}\epsilon = \frac{\epsilon}{2} <1.$$
So then we have left a set of "length" (really, "measure") greater than $1-\displaystyle \frac\epsilon2 $.
You should see that it is immediate (use the same idea I just gave) that a countable set can be covered by a countable collection of intervals of total length less than any positive number. Hence our set cannot be countable.
By construction, our set has no rationals.
Note: that we dealt with the rationals was inconsequential. The only property of the rationals that we needed for this construction was the fact that they are countable.
