Let
$ A= \{ f \vert f: \mathbb{Z}^{+} \rightarrow \{ 0,1,2,3,4,5,6,7,8,9 \} \}$.
Now, if we want to prove that it is uncountable, we can prove that there doesn't exist a bijection between $\mathbb{N}$ and $A$.
Let's assume that there exists a bijection $\Phi : \mathbb{N} \rightarrow A, n \in \mathbb{N}$, such that $ \Phi(n) \rightarrow f_n$, so that among $f_n$, exist all functions in $A$. Let's define a function
$g : \mathbb{Z}^{+} \rightarrow \{ 0,1,2,3,4,5,6,7,8,9 \} $ in the following way:
$$g(x)=\begin{cases} 0, & f(x)=1 \\ 1, & f(x)=2 \\ 2, & f(x)=1 \\ ... \end{cases}$$
for all $x \in \mathbb{Z}^{+}$
I didn't go all the way from $0$ to $9$ because either way I think it's sufficient to show that at least one value is different.
It's evident that such a function $g(x)$ can't be contained among the functions $f_n, n \in \mathbb{N}$ since its value is different from all values of $f_n, n \in \mathbb{N}$ for at least one $n$.
My question is: Is this a valid proof? I saw a similar approach to showing that the set of all functions from R to R has a bigger cardinality than $c$, and I used an analogous approach but I'm so far not convinced that this is correct.