Counterexample:

For $a\in\mathbb Q$ let $L_a=\{x\in\mathbb Q:x\lt a\}$, $R_a=\{x\in\mathbb Q:x\gt a\}$.

Let $H=(\mathbb Q,E)$ where $E=\{L_a:a\in\mathbb Q\}\cup\{R_a:a\in\mathbb Q\}$.

Clearly $H$ is a $T_1$-hypergraph. Consider any $E_1\subseteq E$ such that $(\mathbb Q,E_1)$ is $T_1$. Choose $a\in\mathbb Q$ so that $L_a\in E_1$. Let $E_2=E_1\setminus\{L_a\}$, a proper subset of $E_1$. I claim that $E_2$ is $T_1$.

Assume for a contradiction that $E_2$ is not $T_1$. Then there exist $x,y\in\mathbb Q$ such that $x\ne y$ and there is no $e\in E_2$ such that $e\cap\{x,y\}=\{x\}$. Since $E_1$ is $T_1$, we must have $L_a\cap\{x,y\}=\{x\}$, i.e., $x\lt a\le y$. Choose $z\in\mathbb Q$ so that $x\lt z\lt a$. Since $E_1$ is $T_1$, there is some $L_b\in E_1$ such that $L_b\cap\{x,z\}=\{x\}$, *i.e.*, $x\lt b\le z\lt a\le y$. But then $L_b\in E_2$ and $L_b\cap\{x,y\}=\{x\}$, a contradiction.

**P.S.** Here is an example where the edges of the hypergraph are finite sets: $H=(V,E)$ where $V=\omega=\{0,1,2,3,\dots\}$ and
$$E=\{\{0,1\},\{0,2\},\{0,3\},\{0,4\},\dots\}\cup\{\{1,2\},\{1,2,3\},\{1,2,3,4\},\dots\}.$$