Let $X$ be a topological space. $A\subseteq X$ is connected if and only if for any disjoint, open sets $U,V\subseteq X$, we have $A\subseteq U\cup V$ implies $A\subseteq U$ or $A\subseteq V$.
Forward: Suppose $A$ is connected. Suppose $A\subseteq U\cup V$ for disjoint, open sets, $U,V$. Then, $A=(U\cup V)\cap A =(U\cap A) \cup (V\cap A)$. Since $A$ is connected, $U\cap A=\varnothing$ or $V\cap A=\varnothing$ If $U\cap A = \varnothing$ then since, $U\cap V=\varnothing$ , we have $A\subseteq V$.
Backwards: I suppose $A$ is disconnected. So, $A$ can be expressed as the disjoint union, of two non-empty open sets, $U_A,V_A$ i.e. $A=U_A\cup V_A$ In particular, $U_A=U\cap A$ and $V_A=V\cap A$ where $U, V$ are open in $X$. So $A\subseteq U\cup V$.
However, the issue is that $U\cap V$ may not be empty. I'm not sure how to proceed from this stage.