If you do not like Cantor's Diagnol Argument, possibly because the same real number might not be unique, or the fact that rationals have repeating decimals. Then consider, a another variation of the Diagnol Argument.

Define $\displaystyle \mathbb{I}=\{0<x<1|x\in\mathbb{R}, x \not \in \mathbb{Q}\}$. Thus this set is the set of all irrationals. Next, by the theory of continued fractions, every irrational number can be expressed as a unique infinite continued fraction. By that you assume that the irrationals are countable thus you can list them for example

$\displaystyle [0;2,4,5,1,...$

$\displaystyle [0;8,2,1,1,...$

$\displaystyle [0;5,2,7,1,...$

$\displaystyle [0;1,2,3,4,...$

.....

Now employ the diagnol argument and you demonstrated that $\displaystyle \mathbb{I}$ is uncountable.

Switch from number theory to set theory:

we have that,

$\displaystyle |\mathbb{I}|>\aleph_0\mbox{ and } |\mathbb{Q}|=\aleph_0$Thus,

$\displaystyle \mathbb{I}\cup\mathbb{Q}>\aleph_0$.

Thus,

$\displaystyle \{0<x<1|x\in\mathbb{R}\}$ is uncountable.

Q.E.D.