Not any Cantor like set with irrational endpoints will work. For example, if you restrict the length of removed segment to $\epsilon^n$ at each state at the midpoint of the interval, you will run into issues of not reducing the measure to 0 and will have rationals left over (there will be interval near $a$). (even though the method works if you put more restriction).
Just use the Cantor set. Let $C$ be the 1/3 Cantor set (removing 1/3 each time). It's uncountable, compact, and all element are rational or transcendental (I will refer to wikipedia's Cantor set page. Please look it up with more detail).
Let set $S = \frac{\sqrt{2}}{2}C$. You can tell that $S \subset [0,1]$, also every element in $C$ that is rational will become irrational, transcendental numbers are not algebric number, so if you multiply by $\frac{\sqrt{2}}{2}$ which is a algebric number, you still get a non-algebric number which can't be rational. It's is rather clear it's uncountable (there is a bijection from $S$ to $C$. (It is also compact).