real analysis - Proving that R is uncountable

Is the following proof for the uncountability of $\Bbb{R}$ sufficient? We first assume that the interval $(0,1)$ is countable. So we can define a bijection $f:\Bbb{N}\rightarrow(0,1)$

$$x_{1}=x_{11}x_{12}x_{13}\\x_{2}=x_{21}x_{22}x_{23}\\x_3=x_{31}x_{32}x_{33}\\.\\.\\.$$

Where $x_{ij}$ is the digit in the $jth$ decimal place of the $ith$ number in the list. Now construct some number $y$ whose $jth$ decimal place $y_{j}=x_{ii}+1$ when $x_{ii}\neq9$ and $0$ otherwise. But $y$, while clearly in $(0,1)$, is not in the list, for it differs from $x_{1}$ in the first decimal, $x_2$ in the second, and so on. So $f:\Bbb{N}\rightarrow(0,1)$ is not surjective, and so not a bijection. $(0,1)$ is therefore not countable, and so neither is $\Bbb{R}$.