Let be $f: \mathbb{N}\rightarrow{}\mathbb{N}$ a function. I need to prove that exist a continuous function $g: \beta\mathbb{N}\rightarrow{} \beta\mathbb{N}$ such that $g↾\mathbb{N}=f$.
I have thought in the function $g: \beta\mathbb{N}\rightarrow{} \beta\mathbb{N}$, $g(U)=\left\{{X\subseteq\mathbb{N}: \exists{} A \in{} U \text{ such that } f(A)\subseteq X}\right\}$, but I could not prove that $g↾\mathbb{N}=f$.
Your choice of $g$ is fine. Let $n\in\Bbb N$. Recall that $n$ is identified with the principal ultrafilter $p_n=\{A\subseteq\Bbb N:n\in A\}$. Thus,
$$\begin{align*} g(p_n)&=\{X\subseteq\Bbb N:f[A]\subseteq X\text{ for some }A\in p_n\}\\ &=\{X\subseteq\Bbb N:f[A]\subseteq X\text{ for some }A\subseteq\Bbb N\text{ such that }n\in A\}\\ &=\{X\subseteq\Bbb N:f(n)\in X\}\\ &=p_{f(n)}\;, \end{align*}$$
which we identify with $f(n)$. Thus, with the usual identifications we have $g\upharpoonright\Bbb N=f$.
For continuity and more information about function extensions see this answer.
