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I am studying measure theory and looking for how to prove this. I would like to achieve this without the use of measure theory.

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    Use that the union of a finite or countable collection of countable sets is countable. So if the union is uncountable ... – 2017-01-31
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    Suppose to the contrary that $[0, 1] \setminus \mathbb{Q}$ is countable, where $f : \mathbb{N} \to [0, 1] \setminus \mathbb{Q}$ is a surjection, and let $g: \mathbb{N} \to \mathbb{Q}$ be another surjection (which we know exists). Then set $h(2k - 1) = f(k), h(2k) = g(k)$. We thus get a surjection $h : \mathbb{N} \to \left( [0, 1] \setminus \mathbb{Q} \right) \cup \mathbb{Q}$, i.e. $[0, 1]$ is countable. – 2017-01-31
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    If $[0,1]\setminus\mathbb{Q}$ were countable, $[0,1]$ would be countable too (the union of two countable sets is countable), but $[0,1]$ is not countable by Cantor's diagonal argument. – 2017-01-31

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First, we show that $\mathbb{Q}$ is countable and that the interval $[0,1]$ is uncountable.

Once we have shown these facts, then we assume that $[0, 1] \setminus \mathbb{Q}$ is countable.

But then $$[0, 1] = \left( [0, 1] \cap \mathbb{Q} \right) \cup \left( [0, 1] \setminus \mathbb{Q} \right).$$

Finally, we show that the union of any two (or finitely many!) countable sets is countable, and so is any (infinite) subset of a countable set, thus arriving at a contradiction.

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    Sums it up nicely~ – 2017-01-31
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    Well, $[0,1] = (\mathbb{Q}\cap [0,1]) \cup ([0,1]\setminus \mathbb{Q})$. $2$ certainly isn't in $[0,1]$. – 2017-01-31
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We can find an uncountable subset of $[0,1]\backslash \mathbb{Q}$. Here's a suggestion. Collect all numbers with specified decimal expansion: For any $0,1$ sequence $a=(a_n)$, define

$$ b_{a}=0.b_1 b_2 b_3 \cdots b_n \cdots, $$ where $b_i=0$ if $i\neq n!$, $i\neq n!+1$, $n\geq 2$ and $b_{n!}=1$, $b_{n!+1}=a_n$, $n\geq 2$.

Collection of all such numbers is uncountable because it contains a copy of $\{0,1\}^{\mathbb{N}}$.

Added: This set also provides an example of a perfect set consisted entirely of irrational numbers.