Cantor's theorem: the reals are uncountable

Suppose, for contradiction, that $[0,1]$ were countable: every real number in it appears exactly once in some enumeration $x_1, x_2, x_3, \ldots$. Write each in decimal form as $x_n = 0.d_{n1}d_{n2}d_{n3}\ldots$, choosing the expansion that does not end in an infinite string of 9's when a number has two representations.

Now build a new number $y$ digit by digit, one digit at a time, by looking down the diagonal of this list: $y = 0.e_1e_2e_3\ldots$, where the $n$-th digit is defined by $e_n=\begin{cases}5 & d_{nn}\neq5\\6 & d_{nn}=5\end{cases}$. Restricting the choice to $\{5,6\}$ guarantees $y$ never ends in all 0's or all 9's, so its decimal expansion is unique and unambiguous.

For every index $n$, the number $y$ differs from $x_n$ at the $n$-th decimal digit by construction ($e_n\neq d_{nn}$), so $y\neq x_n\ \forall n$. Since $y$'s digits are all $5$ or $6$, $y\in[0,1]$.

But then $y$ is a real number in $[0,1]$ that is not equal to any $x_n$ in the supposedly complete list — a contradiction, since the list was assumed to contain every element of $[0,1]$. Therefore no enumeration of $[0,1]$ can exist, and $[0,1]$ (hence the larger set $\mathbb{R}$) is uncountable.