I was shown this statement to be a definition. However, I think that the actual statement would be.
A set is open iff its complement is closed.
This way it's a biconditional and it's also true- this is my proof, with C being a continuum that is nonempty, has no first or last point and has an ordering (<): Suppose $U$ is open. Let $x$ be a limit point of $C \setminus U$. It should follow that if $C \setminus U$ is closed, then $x \in (C \setminus U)$. Now we know that $\forall$ regions $R$ containing $x$, there is a nonempty intersection with $C \setminus U$. Also, $x$ cannot be a point of the interior of $U$ because any intersection with a region containing it would be empty, which wouldn't be in accordance to our assumption that $x$ is a limit point of $C \setminus U$. So $x$ is not in the interior of $U$ or $x \notin U$. This would mean that $x \in (C \setminus U)$ and thus $C \setminus U$ is closed.
I was wondering if this proof was reasonable, if there were any gaps in logic, and if this modification of the definition I presented was viable and justified.