I have some troubles with the next exercise. Really, I don't know how can I do it.
Let $X$ be a discrete space and $U\subseteq\beta X$ an open set. Then, $\text{cl}_{\beta X}(U)$ is an open set.
I thought in the charasteristic function over $U\cap X$, i.e., $\chi_{U\cap X}:X\to\mathbb{R}$ and, then, take $f$ the continuous extension of $\chi$ to $\beta X$. I think that $f=\chi_{\text{cl}(U)}$. But, really I don't know if it is correct. Any hint? Thanks in advance.
Yes, you reasoning are correct. Indeed, the set $f^{-1}(1)$ is a closed subset of $\beta X$ and $f^{-1}(1)\supset U$, so $f^{-1}(1)\supset \operatorname{cl} U$. On the other hand, the function $f$ equals $0$ on a dense subset $X\setminus U$ of a set $\beta X\setminus \operatorname{cl} U$, so $f(\beta X\setminus \operatorname{cl} U)=\{0\}$.
PS. Even more general results holds, as show the following quotations from Engelkng's "General topology".
I recall that a topological space $X$ is called extremally disconnected if $X$ is a Hausdorff space and for every open set $U\subset X$ the closure $\overline{U}$ is open in $X$.
I recall that two subsets $A$ and $B$ of a topological space are called completely separated if there exists a continuous function $f:X\to [0,1]$ such that $f(x)=0$ for each $x\in A$ and $f(x)=1$ for each $x\in B$.