Let A be a set of all infinite sequences consisting of 0's and 1's. Prove that A is not countable. Sequences such as {010101010101...., 10100100100...., etc}

if i am not wrong these sequences can represent all the real numbers in the binary format, so a such a set will not be countable. but i am not sure how to go about proving it and how  cantor's diagonalization fits in this case.
Assume it is countable, i.e. there exists an enumeration $(a_n)$ of $A$.
Let $a_i:=a_i^1\cdots a_i^n\cdots$, $i=1,2,\cdots$. The element 
$$b=b_1\cdots b_n\cdots$$
such that $b_i\neq a_i^i$ does not coincide with any $a_i$.
Suppose $\{0,1\}^{\mathbb{N}}$ is countable. Then we can write $\{0,1\}^{\mathbb{N}}$ as a sequence which elements are sequences of $0$s and $1$s,  $(x^1, x^2, \dots ) $. Because this has every sequence of $0$s and $1$s in fact we should have the sequence $(y_n)_{n\geq 1}$ such that $y_k = 1 -x_k^k$ that is $1$ minus the $k$-th element of the $k$-th sequence (note that this is well definided, that is $y_n$ is a sequence of $0$s and $1$s), now it follows that $(y_n)_{n\geq 1}$  must be an element of $(x^1, x^2, \dots ) $ for some $j$ that is $(y_n) = (x^j_n)$, we have arrived at a contradiction because:

$$y_j = x_j^j \quad \text{  and } \quad y_j = 1 - x_j^j$$