Is $\mathcal{P}(\omega)/fin$ with the interval topology a connected space? (You find the definition of $\mathcal{P}(\omega)/fin$ here.)
Remark: According to this, the interval topology of $\mathcal{P}(\omega)/fin$ is not Hausdorff, but I haven't found out whether it contains non-trivial clopen sets.
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.
