Show the set of all functions from $\mathbb{N}$ to $\{0, 1\}$ is uncountable using a contradiction. This is what I have written: 
By contradiction, assume it is countable. Write $S=\{\text{all functions } \mathbb{N} \rightarrow \{0,1\} \}$. Then, we can find a bijection $\mathcal{H}: S \rightarrow \mathbb{N}$. Now, I would like to check how to incorporate Cantor's method to find the contradiction. Would it be right to think of each function as a binary representation (because they map to either $0$ or $1$)? So, I will write 
$f(1) \mapsto a_{11}a_{12}a_{13}...$
$f(2) \mapsto a_{21}a_{22}a_{23}...$
$f(3) \mapsto a_{31}a_{32}a_{33}...$
where $a_{ij} \in \{0,1\}$.
and so on. So for example, $f(1)$ has input any natural number, so it will spit out either a $0$ or a $1$, and I have written all possibilities in a list. 
Then, I define a function in $S$ that is $0$ if a string value is $1$ and $1$ if the string value is $0$. 
I have one more question: what is the meant by the notation $\{0,1\}^{\mathbb{N}}$?
Thank you.

Your idea is correct, Using the binary representation actually makes the explanation way easier. You can always get a binary number that is not in the list and obtain a contradiction using cantor's diagonal method
