(I'm not normally dealing much with set theory and logic, so excuse me if my choice of words below seems a bit off the traditional terminology.) Where does the following Cantor-inspired argument go wrong?
Let $\Sigma = \{A_0,A_1,A_2,\ldots\}$ be the countable collection of statements with one free variable such that for all $n\ge 0$, $$ \exists!x_n\in [0,1):A_n(x), $$ where by $[0,1)$ I mean the interval. ($\Sigma$ is countable, right? Statements must consist of finitely many characters after all.) Now write $x_n$ in decimals as $$ x_n =\sum_{k=1}^\infty x_n^{(k)}10^{-k} $$ where $x_n^{(k)}\in\{0,1,\ldots,9\}$ for all $k\ge 1$ (we make sure never to write e.g. $0.1$ as $0.099999\ldots$). Now put $$ y^{(k)}=\begin{cases} 1 & \text{if $x_k^{(k)}\neq 1$}\\ 2 & \text{else}. \end{cases} $$ Unsurprisingly, we put $$ y:=\sum_{k=1}^\infty y^{(k)}10^{-k}. $$ Now $y\neq x_n$ for all $n$, just like in Cantor's classical argument. But the above definition of $y\in [0,1)$ should (I think---but this could be where I'm wrong) be possible to formalise in first-order logic. In other words, there exists a statement $A$ such that $y$ is the one and only number in $[0,1)$ such that $A(y)$ is true. Thus $A\in\Sigma$, which is obviously a contradiction.