Let $\kappa$ be a limit ordinal. Taken from the definition of a closed unbounded set, we say a subset $C\subseteq\kappa$ is closed in $\kappa$ if and only if
$\sup(C\cap\alpha)=\alpha<\kappa\implies\alpha\in C$
Is this definition equivalent to being closed with respect to the order topology on $\kappa$? Wikipedia claims it should be, but I cannot prove it.
The order topology has as its sub-base all upper and lower sets, $U(\alpha) = \{\beta: \beta > \alpha\}$ and $L(\alpha) = \{\beta: \beta < \alpha\}$.
This implies that in an ordinal $\kappa$, all successor ordinals $\alpha+1\in \kappa$ are isolated points, because $\{\alpha+1\}= (\alpha, \alpha+2)=U(\alpha\} \cap L(\alpha+2)$. At limit ordinals $\beta\in \kappa$ we have that a local base is all sets of the form $(\alpha, \beta]$ (this is open as $U(\alpha)\cap L(\beta+1)$), so at limit ordinals the topology "looks to the left".
Now suppose that $C$ is order-closed and that $\sup(C\cap \alpha) = \alpha < \kappa$ and $\alpha$ is a limit. Then if $(\gamma, \beta]$ is any basic open neighbourhood of $\alpha$, with some $\gamma < \alpha$, we note that $\gamma$ is no upperbound for $C\cap \alpha$ and so some $\gamma'\in C\cap \alpha$ exists with $\gamma' > \gamma$. This means that $\gamma' \in (\gamma, \alpha]$ so that we know that every basic neighbourhood of $\alpha$ intersects $C$, so $\alpha \in \overline{C}=C$, as required.
A similar argument shows that if $C$ obeys the "$\sup$-property", it is order closed: let $\alpha \in \overline{C}$. If $\alpha$ is isolated it has a neighbourhood $\{\alpha\}$ that must intersect $C$, so $\alpha \in C$. So now assume $\alpha$ is a limit. $\alpha \in \overline{C}$ implies that for every $\gamma < \alpha$ the basic neighbourhood $(\gamma, \alpha]$ intersects $C$. This readily implies that $\alpha = \sup(C \cap \alpha)$ (as $\alpha$ is surely an upperbound for $C \cap \alpha$ and the previous remarks shows no smaller one can exist. The property then implies $\alpha \in C$ and so $\overline{C} \subseteq C$ and $C$ is order-closed.