Asking for Various proofs of uncountability of $[0,1]$ How many different proofs are there of the uncountability of the set $[0,1]$ ? I know of the nested interval proof and the Baire Category theorem proof ; please suggest other proofs . Thanks in advance . 
$EDIT$ : Is there a proof of this uncountabilty using Shroder-Bernstein Theorem ?
 A: Matthew H. Baker of Georgia Institute of Technology uses a game.

A: The game proof given by BCLC is nice!
The classical ones that I know are the following:


*

*Direct use of the nested interval theorem : given a sequence of real numbers, construct a real number different from all the numbers in the sequence as the common point of a nested sequence of intervals. (This proof is already pointed out in the question).

*The proof using the decimal expansion of real numbers : given a sequence of real numbers given by their decimal expansion, use the diagonal method to produce a real number whose decimal expansion is different from all the ones you are given. (This one should be that suggested in Prahlad's comment).

*A variant of 2, where one first shows that there are at least as many real numbers as subsets of the integers (for example, by constructing explicitely a one-to-one map from $\{ 0,1\}^{\mathbb N}$ into $\mathbb R$), and then show that $\mathcal P(\mathbb N)$ is uncountable by the method you like best.

*The Baire category proof : $\mathbb R$ is uncountable because 1-point sets are closed sets with empty interior. (Also pointed out in the question).

*The measure proof : $\mathbb R$ is uncountable because $1$-point sets have measure $0$. (This is in idm's answer).
5'. The measure proof without measure. First show that a countable set $D$ is negligible in the following sense: for any $\varepsilon >0$, $D$ can be covered by a sequence of intervals $(I_n)$ such that $\sum_0^\infty \vert I_n\vert<\varepsilon$. Then show that a nontrivial interval $I$ is not negligible in this sense, by proving that if $I$ is covered by a sequence of intervals $(I_n)$, then $\sum_0^\infty \vert I_n\vert\geq \vert I\vert$ (this requires some work).
Without any doubt, there are lots of other proofs. For example, an amusing one in this paper: http://arxiv.org/pdf/0901.0446v1.pdf I also found a proof using "only" the completeness property in Theorem 8.1 of this paper: http://arxiv.org/pdf/1209.5119.pdf
A: There are many I think. 
I can give you this one:
Let $m$ be Lebesgue measure. If a set is countable, then its measure is null (but the converse is not true e.g. the cantor set). We have that $$m([0,1])=1>0.$$ Therefore, [0,1] is uncountable.
