I know that this proof may sound ridiculous, but I'm really curious to find out if it's logically correct(and whether there are some circularities). Since $|[0,1]|=|\mathbb{R}|$, we have to simply show that $[0,1]$ is not countable. This theorem holds:
$f:[a,b]\to \mathbb{R} \\ \text{Let } D_f \text{ be the set of discontinuities of }f \\ |D_f|\leq\aleph_0 \Rightarrow f \text{ is Riemann-integrable in }[a,b] $
Now I can define the Dirichlet function in $[0,1]$: $\chi(x)=\begin{cases} 1 \ \ \text{if }x\in\mathbb{Q} \\ 0 \ \ \text{otherwise} \end{cases}$
Clearly $D_\chi=[0,1]$. By contradiction, if $|[0,1]|=\aleph_0$, then it would mean that $\chi$ is Riemann-integrable in $[0,1]$, but it's not: this is a contadiction!
So $|[0,1]|>\aleph_0$.
It is correct and as far as I can see this does not contain circularities. Perhaps it is a little cleaner to use contraposition instead of contradiction though.
FYI, a more general statement than the one you described holds. Let $f: [a,b]\to \mathbb{R}$ be a bounded function. Then $f$ is Riemann-integrable if and only if $D_f$ has Lebesgue measure $0$. Of course, if $D_f$ is countable then it has Lebesgue measure $0$ since singeltons don't have a length.