The answer is yes, because any two nonempty open sets have points
in common. And this also shows directly that the space is not
Hausdorff.

From what you describe in the other question, the topology is
generated by the complements of the upper cones $\uparrow
f=[f,1]=\{h\mid h\geq^* f\}$ and the lower cones $\downarrow
g=[0,g]=\{h\mid h\leq^* g\}$, where I am regarding the points as
equivalence classes of functions $f:\omega\to 2$ (but I suppress
the equivalence classes). Thus, a set is open if it is a union of
finite intersections of such cone complements. Let us take those
finite intersections as basic open sets.

What I claim is that any two nonempty basic open sets have
nonempty intersection.

Suppose that $U$ is the intersection of the complements of the
intervals $[f_i,1]$ and $[0,g_j]$ for finitely many $i,j$. So a
function $h:\omega\to 2$ is in $U$ just in case it is not
almost-above any $f_i$ and not almost-below any $g_j$.

Similarly, suppose $V$ is the intersection of the complements of
the intervals $[f_i',1]$ and $[0,g_j']$. So $V$ consists of the
functions $h$ that are not almost-above any $f_i'$ and not
almost-below any $g_j'$.

Assume that $U$ and $V$ are not empty. Thus, we may assume that
the functions $f_i$ are not almost-always $0$ and the $g_j$ are
not almost-always $1$. It follows that there are infinite
sets $A_i$ and $B_j$ such that $f_i(n)=1$ for $n\in A_i$ and
$g_j(n)=0$ for $n\in B_j$. And similarly $A_i'$ and $B_j'$ for
$f_i'$ and $g_j'$. By shrinking these sets, we may assume that all
$A_i, A_i', B_j, B_j'$ are pairwise disjoint.

Let $h$ be the function that is $0$ on every $A_i$ and $A_i'$ and
$1$ on every $B_j$ and $B_j'$. Thus, $h$ is not above any $f_i$
nor any $f_i'$ and not below any $g_j$ nor $g_j'$. So $h\in U\cap
V$, as desired.

2more comments