If $(P,\leq)$ is a poset and $x\in X$, we let $\downarrow x = \{p\in P: p \leq x\}$, and $\uparrow x$ is defined dually. The collection $$\Big\{P\setminus (\downarrow x): x\in P\Big\} \cup \Big\{P\setminus (\uparrow x): x\in P\Big\}$$ is a subbasis for the interval topology $\tau_i(P)$ on $(P,\leq)$.
Question. If we endow the poset ${\cal P}(\omega)/(\text{fin})$ with the interval topology, is the resulting space connected, or even path connected?
The space is hyperconnected and thus also connected. Indeed, it suffices to show that, for any $x_1, \cdots, x_n \in \mathcal{P}(\omega)/(\text{fin})$ which are not $[\omega]$, and for any $y_1, \cdots, y_m \in \mathcal{P}(\omega)/(\text{fin})$ which are not $[\varnothing]$, we have,
$$\bigcup_{i=1}^n (\downarrow x_i) \cup \bigcup_{j=1}^m (\uparrow y_j) \neq \mathcal{P}(\omega)/(\text{fin})$$
Indeed, pick $X_i, Y_j \subset \omega$ as represenatives of $x_i, y_j$, resp. Since $X_i$ are co-infinite, for each $1 \leq i \leq n$, we may choose, $A_i \subset X_i^c$ infinite. By removing redundant elements through the following procedure, we may assume $A_i$ are pairwise disjoint: For $i_1 \neq i_2$, if $A_{i_1} \cap A_{i_2}$ is co-infinite in both $A_{i_1}$ and $A_{i_2}$, remove $A_{i_1} \cap A_{i_2}$ from both sets. Otherwise, by removing finitely many elements, one set is contained in the other. WLOG, we may assume $A_{i_1} \subset A_{i_2}$. Then take an infinite co-infinite $B \subset A_{i_1}$, change $A_{i_1}$ to $B$, and change $A_{i_2}$ to $A_{i_2} \setminus B$.
Let $A = \bigsqcup_{i=1}^n A_i$. We inductively remove elements from $A$ to ensure $[A]$ is not in $\uparrow y_j$ for any $j$: At step $k$, assume we have $A^k = \bigsqcup_{i=1}^n A_i^k$ where $A_i^k \subset X_i^c$ is infinite for all $i$ and furthermore $A^k$ does not contain $Y_j$ up to finite sets, for all $j < k$. Now, consider $Y_k$. If $A^k$ does not contain $Y_k$ up to finite sets, simply set $A^{k+1} = A^k$. Otherwise, since $Y_k$ is infinite and is contained in $A^k = \bigsqcup_{i=1}^n A_i^k$ up to finite sets, it must be the case that $A_{i_0}^k \cap Y_k$ is infinite for some $i_0$. Pick an infinite co-infinite subset $B \subset A_{i_0}^k \cap Y_k$. Then set $A_i^{k+1} = A_i^k$ for all $i \neq i_0$ and $A_{i_0}^{k+1} = A_{i_0}^k \setminus B$.
At the end, $A^{m+1}$ contains infinitely many elements from $X_i^c$ for all $i$, so $[A^{m+1}] \notin \; \downarrow x_i$ for all $i$. Moreover, the inductive procedure ensures that $[A^{m+1}] \notin \; \uparrow y_j$ for all $j$. Thus,
$$[A^{m+1}] \notin \bigcup_{i=1}^n (\downarrow x_i) \cup \bigcup_{j=1}^m (\uparrow y_j)$$
So, the set on the RHS is indeed not the entirety of $\mathcal{P}(\omega)/(\text{fin})$, as desired.
