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 $\Bbb{R}$.
The problem with this proof is that it assumes there is a one-to-one mapping between the real numbers in $(0,1)$ and the numbers that can be represented as $0.x_1x_2x_3\ldots$ 
But this is wrong. Let's say $x_1 = \frac{1}{2},$ but you choose the unusual representation $0.49999\ldots$ for it. Now it happens by accident that $x_{22}=x_{33}=x_{44}=\ldots=9$ (you have not excluded this case from consideration, and you can obviously find a sequence of different real numbers $x_2,x_3,\ldots$ with this property.) Then your $y$ will be $0.50000$ which represents exactly the same real number as $x_1$, and you haven't proven anything.

Sure, all those drawbacks originating from the usage of the decimal representation of real numbers can be addressed somehow. But there are much simpler proofs which do not have to deal with the ambiguity of decimal representation.

If you want to stay with the "diagonal argument", you have to ensure at least that the $y$ cannot be a different decimal representation of a real number that is already in the list.
You can simplify the diagonal argument considerably by considering the binary representation of real numbers. Then you simply go along the diagonal flipping 0s to 1s and 1s to 0s.

The binary version goes like this: 

Consider all binary sequences. Each of these sequences maps to a number in $[0,1]$ via:

$$\{a_n\} \mapsto \sum_{j=1}^{\infty} \frac{a_j}{2^j}.$$

Note that because of the geometric series result, this sum is in $[0,1]$, and hence we have mapped each of these sequences to a number in $[0,1]$. Now suppose the set of these sequences were countable. Then we could simply order the sequences. But, if we can do this, then I can create a new sequence where the $j$th entry is the negation of the $j$th sequence's $j$th entry. By construction, this sequence is different from all the others. This contradicts the assumption that this set of sequences is countable. Hence, $[0,1]$ must be uncountable, and therefore $\mathbb{R}$ must be uncountable.