Let $X$ be $T_1$ are equivalent: $X$ is $T_1$ and every subset of $X$ is intersection of open set $U$ such that $x\in U$.
Then if $X$ is $T_1 \Longrightarrow \{x\}=\bigcap\limits_{U\in\mathcal{E}(x)}U$ ($x$ subset of $X$)
On the other hand, if $\{x\}=\bigcap\limits_{U\in\mathcal{E}(x)}U$ . Let $y\in X$ $(x\neq y)$ .For $\{x\}=A$ it's clear that exist a neighborhood $V\in\mathcal{E}(x)$ such that $y\notin V$. Same if $A=\{y\}$. In conclusion $X$ is $T_1$
For $\implies$, it seems like you're just stating the statement, without proving anything. Regardless, this is easy to prove: For any $y \in X$ were $y \neq x$, there exists $V \ni x$ open such that $y \notin V$. Since $\bigcap_{U \in \mathcal{E}(x)} U \subseteq V$, $y \notin \bigcap_{U \in \mathcal{E}(x)} U$, hence $\{x\} = \bigcap_{U \in \mathcal{E}(x)} U$.
For $\impliedby$, your idea is fine but you can phrase it better. Here is a rewritten version of your proof: Since $y \notin \bigcap_{U \in \mathcal{E}(x)} U$, there exists $U \in \mathcal{E}(x)$ such that $y \notin U$, i.e. $x \in U$ but $y \notin U$. Since $x \in X$ is arbitrary, $X$ must be $T_1$.