Are there any different proof of the uncountability of $[0, 1]$?

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

1. $\alpha < A_1 < B_1 < \beta$.
2. 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

1. $\alpha < A_1 < A_2 < \cdots < A_r < B_r < \cdots < B_2 < B_1 < \beta $.
2. 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

1. Über unendliche, lineare Punktmannigfaltigkeiten
2. The Calculus Gallery