Statement
If $X$ is a $T_1$ space and if $X$ is not discrete then there exist $x_0\in X$ that has a local basis $\mathcal{B}(x_0)$ such that $\bigcap\mathcal{B}(x_0)$ is not open.
Proof. So let be $X$ a $T_1$ space that is not discrete and we suppose that any local basis $\mathcal{B}(x)$ of $x\in X$ is such that $\bigcap\mathcal{B}(x)$ is open. So if $X$ is not discrete then there exist $x_0\in X$ such that $\{x_0\}$ is not open and so this means that $\bigcap\mathcal{B}(x_0)\neq\{x_0\}$, that is there exist $y\in X$ such that $y\in\bigcap\mathcal{B}(x_0)$ and so this means that the nhood sistem of $x$ is the same of $y$ but this is impossible if $X$ is $T_1$, because indeed otherwise this would mean that if $U$ is open and $x\in U$ then $y\in U$.
So is the proof correct? if not, how to prove the statement? Could someone help me, please?
Your proof is almost correct but you seem to assume that $\cap \mathcal{B}(x_0)$ is open, which there is no reason for it to be.
However, let's just argue that $\cap \mathcal{B}(x_0)=\{x_0\}$ for some suitable neighbourhood basis of $x_0$ for any $x_0$ in any $T_1$ space $X$. Indeed, let $\mathcal{B}$ be the set of all open $U$ such that $x_0\in U$. Then, given $x_0 \neq y\in X,$ we just want to argue that $y\not \in \cap \mathcal{B}(x_0)$.
As you said, since $X$ is $T_1$, there is $U\in \mathcal{B}(x_0)$ such that $y\not \in U$. Hence, since $ \cap \mathcal{B}(x_0)\subseteq U$, we conclude that $y\not \in \cap \mathcal{B}(x_0)$. Since $y$ was arbitrary, we conclude that $\cap \mathcal{B}(x_0)=\{x_0\}$.
Now, conclude.