Question:
Let $\mathcal{T}, \mathcal{S}$ be two topologies on $X$, and $\mathcal{T}$ is finer than $\mathcal{S}$
If $(X, \mathcal{T})$ is connected, what about $(X, \mathcal{S})$?
Well, if $\mathcal{S}$ is coarser than $\mathcal{T}$, then $\mathcal{S} \subseteq \mathcal{T}$
This means every open set in $\mathcal{S}$ is in $\mathcal{T}$
Then $(X, \mathcal{T})$ is connected if there are no nontrivial, disjoint open sets $A, B \subset X$ such that $A \sqcup B = X$ [Edited per comment]
Since there is no such disjoint open sets $A,B \in \mathcal{T}$, and $\mathcal{S} \subseteq \mathcal{T}$ $\implies$ there is no such open sets $A,B \in \mathcal{S}$. Necessarily $(X, \mathcal{S})$ is connected.
Is my assessment correct?