I was about to answer the same thing as Joel, but here is a more topological perspective. I will show that the lattice of open subsets of $\overline{\mathbb{N}}$ is very similar to the lattice of open subsets of the Stone-Čech compactification $\beta\mathbb{N}$. So, from the localic point of view, there is but a small difference between your space $\overline{\mathbb{N}}$ and $\beta\mathbb{N}$.

View $\beta\mathbb{N}$ as the set of ultrafilters on $\mathbb{N}$ and let $\mathbb{N}^*$ be the remainder $\beta\mathbb{N}\setminus\mathbb{N}$. (Prinicipal ultrafilters are identified with the corresponding point of $\mathbb{N}$, so $\mathbb{N}^*$ is the subspace of nonprincipal ultrafilters.) Recall that the clopen subsets of $\beta\mathbb{N}$ are precisely those of the form

$$\langle A \rangle = \{ \mathcal{U} \in \beta\mathbb{N} : A \in \mathcal{U} \}$$

for $A \subseteq \mathbb{N}$. Note that $A \preceq B$ iff $\langle A \rangle \cap\mathbb{N}^* \subseteq \langle B \rangle\cap\mathbb{N}^*$, so the points of $\partial\mathbb{N}$ can be identified with the clopen subsets $[A] = \langle A \rangle \cap \mathbb{N}^*$ of the remainder $\mathbb{N}^*$ (including the empty set). Given an open set $U \subseteq \beta\mathbb{N}$, the set

$$U' = (U \cap \mathbb{N}) \cup \{ [A] : [A] \subseteq U \}$$

is open in $\overline{\mathbb{N}}$. Conversely, given an open set $V \subseteq \overline{\mathbb{N}}$, your conditions ensure that

$$V' = (V \cap \mathbb{N}) \cup \bigcup \{ [A] : [A] \in V \}$$

is open in $\beta\mathbb{N}$. This correspondence is not perfect since $A \cup B \cup [A] \cup [B] = A \cup B \cup [A \cup B]$, but

$$A \cup B \cup \{[C] : C \preceq A \lor C \preceq B\} \quad\mbox{and}\quad A \cup B \cup \{[C] : C \preceq A \cup B\}$$

are not always the same. However, these are the only errors that occur, i.e. the translation is perfect for open subsets of $\overline{\mathbb{N}}$ whose part in $\partial\mathbb{N}$ is upward directed in the ${\preceq}$ ordering.

Although your space $\overline{\mathbb{N}}$ is interesting, this approximate translation suggests most of its applications could be transferred to work over the well studied space $\beta\mathbb{N}$ instead.

Here is yet another perspective which suggests that there may be more to $\overline{\mathbb{N}}$ after all. The soberification of $\partial\mathbb{N}$, which I will denote $\mathrm{Fil}_{\mathbb{N}}$, is the space of all nonprincipal filters on $\mathbb{N}$, with the topology generated by the basic open sets

$$[A] = \{ \mathcal{F} \in \mathrm{Fil}_{\mathbb{N}} : A \in \mathcal{F} \}.$$

The points of $\partial\mathbb{N}$ can be identified with the filters $\mathcal{F}_A = \{ B : A \preceq B \}$. Note that the space $\mathbb{N}^*$ is also a subspace of $\mathrm{Fil}_{\mathbb{N}}$, which explains the connection found above.

For the pointless topology aficcionados, the space $\mathrm{Fil}_{\mathbb{N}}$ is obtained by imposing the trivial Grothendieck topology on the preorder $(\mathcal{P}\mathbb{N},{\preceq})$ viewed as a category (or, equivalently, the quotient partial order $\mathcal{P}\mathbb{N}/\mathrm{fin}$ as suggested by Joel). The subspace $\mathbb{N}^*$ is similarly obtained by imposing the finite cover (aka coherent) Grothendieck topology on $(\mathcal{P}\mathbb{N},{\preceq})$.

`$\bigcap_{n=0}^\infty \{n,n+1,\ldots\}\cup\partial\mathbb{N} = \partial\mathbb{N}$`

is not open. – François G. Dorais♦ Mar 16 '10 at 21:58`$\partial\mathbb{N}$`

, not the full space`$\overline{\mathbb{N}}$`

. – Harald Hanche-Olsen Mar 16 '10 at 23:04