The answer is yes. Count the rationals in $[0,1]$ as $r_1,r_2,\ldots$, let $I_k$ be an open interval containing $r_k$ of length $3^{-k}$, and let $K=[0,1]\setminus\cup_k I_k$.
This question is somewhat related to the question Perfect set without rationals, but there measure did not come up. For example, the Cantor set-like construction given there by JDH could be made to have positive measure.
I'm not sure if this is the shortest approach, but proving that $\sqrt 2$ is irrational doesn't require too much effort. First, prove that no rational squares to $2$. Then, define the set
$$
E=\{x\in\Bbb Q:x<0\text{ or }x^2<2\} \, .
$$
(Those familiar with the construction of the reals will recognise $E$ as the Dedekind cut of $\sqrt 2$.)
This set does not have a supremum in the rationals: if $r\in\Bbb Q$ is an upper bound of $E$, then $s=(2r+2)/(r+2)$ is a smaller upper bound of $E$. Proof that $s$ is an upper bound of $E$:
$$
\left(\frac{2r+2}{r+2}\right)^2-2=\frac{2(r^2-2)}{(r+2)^2}>0 \, .
$$
Proof that $s<r$:
$$
\frac{2r+2}{r+2}=r-\frac{r^2-2}{r+2}<r \, .
$$
Since the reals have the least upper bound property, and $E$ is obviously non-empty and bounded above, $E$ must have a supremum in $\Bbb R$. And since $E$ does not have a supremum in $\Bbb Q$, this supremum must be irrational.
Remark: this approach doesn't actually prove that $(\sup E)^2=2$, but since we are given the ordered field axioms for free, it is not too much work to prove this. In the comments, user21820's outlines one possible approach. Below, I offer another way of showing that there is a real number which squares to $2$.
Using the least upper bound property of $\mathbb R$, we can prove that if a sequence which is increasing and bounded above, then its supremum is its limit. Now consider the sequence $(a_n)_{n\in\mathbb N}$ given by $a_0=1$, and
$$
a_{n+1}=\frac{2a_n+2}{a_n+2} \, .
$$
It can easily be proven by induction that this sequence is increasing and bounded above. Let $l=\lim_{n\to\infty}a_n$. We have
$$
l=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{2a_n+2}{a_n+2}=\frac{\lim_{n\to\infty}2a_n+2}{\lim_{n\to\infty}a_n+2}=\frac{2l+2}{l+2} \, ,
$$
and so $l^2+2l=2l+2$, meaning that $l^2=2$. Since $(a_n)_{n\in\mathbb N}$ is a positive sequence of terms, we see that $l$ equals the positive square root of $2$.
Best Answer
Here is an alternative to the Cantor set examples, but it requires some measure theory. We can construct an open set of arbitrarily small measure that contains all the rationals by enumerating the rationals and taking the union of open intervals of length $2^{-n}\epsilon$ around the $n$th rational. The complement of this set is then a closed set containing only irrational numbers. If we want a bounded set, just intersect if with a bounded closed interval. As long as that interval's length is larger than $\epsilon$, we're guaranteed to end up with a set of positive measure (and such a set must be uncountable).
On a related note, we can show that if we have an unbounded, uncountable closed set of irrationals then we can construct a bounded set from it. Just take its intersection with each of the unit intervals $[n,n+1]$ for $n\in \mathbb Z$. Since there are only countably many such intervals, at least one of those intersections has to be uncountable.