If $E$ is a subset of $X$, then $E$ is open if and only if the complement $X\backslash E = \{x\in X\mid x\not\in E\}$ is closed.
MY ATTEMPT
Let us prove the implication $(\Rightarrow)$ first.
If $E$ is open, then $E = \text{int}(E)$.
Since $X = \text{int}(E)\cup\partial E\cup\text{ext}(E)$ is partition of $X$, we conclude that
$$E^{c} = \partial E\cup\text{ext}(E) = \partial E^{c}\cup\text{int}(E^{c}) = \overline{E^{c}}$$
which is closed because it contains all its adherent points.
We may tackle the converse implication $(\Leftarrow)$ now.
If $E^{c}$ is closed, then $E^{c}\supseteq \partial E^{c} = \partial E$. Then we have that $\partial E\cap E^{c} = \partial E$. Consequently, we have that \begin{align*} E\cap\partial E = E\cap(\partial E\cap E^{c}) = (E\cap E^{c})\cap\partial E = \varnothing\cap \partial E = \varnothing \Rightarrow E\cap \partial E = \varnothing \end{align*} Thus $E$ is open, and we are done.
Comments on the definitions and results
We say that $E$ is open if $E\cap\partial E = \varnothing$. We do also say that $E$ is closed when $\partial E\subseteq E$.
It also has been proved that $\overline{E} = \text{int}(E)\cup\partial E$.