Denote by $^{S}F$ the set of all functions from set $S$ to set $F$. For example, let $S=\{0, 1\}$ and $F=\mathbb{N}$, then $^{S}F$ denotes the set of all functions from $\{0, 1\}$ to $\mathbb{N}$. This set of all functions can be represented by the set of ordered pairs $(x_0, x_1)$.
Using the notation, I've come up with a proof of existence of a one-to-one correspondence (or a bijection) from $^{\{0\}}\mathbb{N}$ to $^{\mathbb{N}}\mathbb{N}$ as follows:
By using the diagonal argument, we can easily obtain that there's a bijection from $^{\{0\}}\mathbb{N}$ to $^{\{0, 1\}}\mathbb{N}$. Also, by using the same argument we can deduce that there's a bijection from from $^{\{0\}}\mathbb{N}$ to $^{\{0, 1, \dots, k\}}\mathbb{N}, \forall k \in \mathbb{N}$.
Because, this is true for all $k\in\mathbb{N}$, then we can deduce that there's a bijection from $^{\{0\}}\mathbb{N}$ to $^{\mathbb{N}}\mathbb{N}$. To see this, suppose to the contrary that this is not true. Then, there must exist some $k\in\mathbb{N}$ such that there's no bijection from $^{\{0\}}\mathbb{N}$ to $^{\{0, 1,\dots, k\}}\mathbb{N}$. However, this contradicts what we just obtained from the previous paragraph. So, we conclude that there's bijection from $^{\{0\}}\mathbb{N}$ to $^{\mathbb{N}}\mathbb{N}$.
However, the conclusion is not true because this leads to the real number is countable. Still, I can see for myself what's wrong with the above proof?