$X$ has Indiscrete topology $\implies$ any subset of $X$ is connected My attempt: Take any $C \subseteq X$. We show $C$ is connected. Suppose $C = A \cup B$ where $A$ and $B$ are disjoint, nonempty subsets (Open $\in \mathcal{T}_{indiscrete}$). Since only $X$ and $\varnothing$ live in the indiscrete topology, WLOG we can put $A = X$ and $B = \varnothing$. But this implies $C = X$ which is a contradiction. Hence, $C$ is connected. Is this solution correct? any feedback? thanks

You have the basic idea, but you’re missing some crucial pieces. If $C$ is not connected, you can’t guarantee that your $A$ and $B$ are open in $X$; you can only guarantee that they are relatively open in $C$. Thus, you want to show that if $A$ and $B$ are disjoint, relatively open subsets of $C$ such that $C=A\cup B$, then one of $A$ and $B$ must be empty.

Since $A$ and $B$ are relatively open in $C$, there must be open sets $U$ and $V$ in $X$ such that $A=U\cap C$ and $B=V\cap C$. But then without loss of generality $U=X$ and $V=\varnothing$, so $B=\varnothing$, as desired and $C$ is connected.

Note that it makes no difference whether $C=X$ or not.