It can be shown that in the Baire space there is a bijection $$\begin{align}\mathcal{N} &\longrightarrow \{f \in {}^\mathcal{N}\mathcal{N} \mid f \text{ is Lipschitz}\}\\ x &\longmapsto \ell_x\end{align}$$ such that the evaluation function $$\begin{align}\mathcal{N}^2 &\longrightarrow \mathcal{N} \\ (x,y) &\longmapsto \ell_x(y)\end{align}$$ is continuous. Here by $\mathcal{N}$ I mean $\omega^\omega$. We can for example construct such a bijection from the standard enumeration of finite sequences of naturals $\omega \rightarrow \omega^{<\omega}$. This bijection gives us a very nice parametrization of Lipschitz functions in the Baire space.
Now on the other hand I've read that there is no parametrization $x\mapsto f_x$ of continuous function s.t. the evaluation map $\mathcal{N}^2 \rightarrow \mathcal{N}, (x,y) \mapsto f_x(y)$ is continuous.
I think that the way to see this is by contradiction, assuming the existence of such parametrization and composing the evaluation function with some continuous function $\mathcal{N} \rightarrow \mathcal{N}^2$
e.g. $$\begin{align}\mathcal{N}&\longrightarrow \mathcal{N}^2& &\longrightarrow \mathcal{N}\\ x &\longmapsto(x,x)& &\longmapsto f_x(x)\end{align}$$
Hence having $\exists y$ s.t. $\forall x (f_x(x) = f_y(x))$ by surjectivity of the parametrization $f_x$.
Now I have tried to use some diagonalization argument to find a contradiction, but without success. Any hint? Is at least the strategy the correct one? Thanks
You are on the right track, but it needs a slight modification. Let me remind you how we applied the diagonalization argument.
How did we prove that there is no surjection from $\omega$ to $\omega^\omega$? Assume that there is a surjection $f$ and consider the function $g$ given by
$$g(n) = f(n)(n)+1.$$
Here the role of $+1$ is important: it makes $g$ different from all of $f(n)$. Your diagonalization lacks a factor corresponding to $+1$, and this is the point we need to modify.
Now examine a new proof. Consider any continuous function $g:\mathcal{N}\to\mathcal{N}$ whose output always differs from the corresponding input (i.e. $g(x)\neq x$ for all $x\in\mathcal{N}$.) An example of these function is $$g(\langle f(0),f(1),f(2),\cdots\rangle)=\langle f(0)+1,f(1)+1,f(2)+1,\cdots\rangle.$$
Now apply the diagonal argument to $g(f_x(x))$. If $f_x$ enumerates all continuous functions, then we can find $y$ such that $g(f_x(x))=f_y(x)$ for any $x$. But it leads to a contradiction.