So, Im trying to show that only the singletons in a set $X$ with discrete topology are connected sets. Pick arbitrary $\{ x \} \subset X$. And suppose $\{ x \} = A \cup B$ where $A$ and $B$ are open relative to $X$ and they are nonempty and disjoint. But, since they are non-empty, $A \cup B$ must contain at least two points, therefore we have a contradiction.

Is this solution correct? Can someone give me some feedback? thanks

Improved Solution: Take any $C \subseteq X$. Suppose $C$ is connected. We show $C$ is a singleton. Since $C$ is connected, then $C$ cannot be written as a disjoint union of elements $A$ and $B$ of $\mathcal{T}\_X$. In other words, if $C = A \cup B \implies C = A $ or $C = B$. Say $C$ contains more than one element. Since $C \in \mathcal{T}\_X$, then $C$ can be written as a union of elements in the basis of $X$, say $C = \bigcup N\_x$. We know $N\_x$, the neighborhoods, are open by definition. If the $N\_x$ are disjoint, then we are done. Otherwise, If $C = A$, then we will end up having $A \cap B \neq \varnothing$ which is a contradiction. Therefore, $C$ must be a singleton.