Are there any different proof of the uncountability of $[0, 1]$? How many ways out there to prove $[0, 1]$ with is uncountable?

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*Proof by the famous Cantor's diagonal argument :

There is no onto function from $\mathbb{N}$ to $[0, 1]$.
Suppose, $f:\mathbb{N} \to [0, 1]$ be any function.
For any $n\in \mathbb{N} $ define
\begin{align}a_n=\Bigg \{ &2 :\text{if the nth decimal place of} f(n)≠2\\
&3 : otherwise\\
\end{align}
$x=\sum_{n\in \mathbb{N}}\frac{a_n}{10^n}$
Then, $x \in [0, 1]$ but $x\notin f(\mathbb{N}) $


*Consider the measure space $(\mathbb{R}, \ell(\mathbb{R}), \lambda) $ , then $[0, 1]$ is Lebesgue Measurable with Lebesgue measure $\ell([0, 1]=1>0$
Hence, $[0, 1]$ is uncountable.


*$[0, 1]$ as a metric subspace of $(R, d_{euclidean}) $ is a complete metric space without any isolated point.

Then by Baire Catagory theorem, $[0, 1]$ must be uncountable.
Are there any different proof of the uncountability of $[0, 1]$?
 A: Here's a sketch of Cantor's original proof that given an interval $(\alpha,\beta)$ and a sequence $\{x_k\}$ of real numbers, there is at least one number $c\in (\alpha,\beta)$ that is not in the sequence.
First find two elements of the sequence $x_{k_1}$ and $x_{k_2}$ with the smallest subscripts such that $x_{k_1} < x_{k_2}$ and $x_{k_1},x_{k_2}\in(\alpha,\beta)$ and denote them $A_1$ and $B_1$. If no such elements exist we are done as we can easily find such a $c$. Then

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*$\alpha < A_1 < B_1 < \beta$.

*If $x_k\in(A_1,B_1)$ we must have $k\geq 3$.

Iterate this process to find sequences $A_r$ and $B_r$ of elements in the sequence $\{x_k\}$ such that

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*$\alpha < A_1 < A_2 < \cdots < A_r < B_r < \cdots < B_2 < B_1 < \beta $.

*If $x_k\in (A_r,B_r)$ then $k \geq 2r +1$.

Again, if this process cannot be continued at any step then we have an interval with at most $1$ point of the sequence and can easily find a number in that interval that isn't among the $x_k$. So we may assume that the process can be continued indefinitely. This leads to an infinite sequence of nested closed intervals
$$[A_1,A_2]\supset [A_2,B_2]\supset\cdots$$
Such a sequence always has at least one point in its intersection, call it $c$. Suppose now that $c=x_N$ for some natural number $N$. Then it must be that $x_N\in(A_N,B_N)$, but by construction this implies
$$N \geq 2N+1$$
which is a contradiction.
Sources

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*Über unendliche, lineare Punktmannigfaltigkeiten

*The Calculus Gallery
