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.
As you wish to prove the backwards implication is true by proving its converse is true, explicitly negating the proposition may help you tackle the problem:
$A$ is disconnected $\implies$ $\exists U,V\subset X$ disjoint open sets s.t. $A\in U \cup V$ and $A\nsubseteq U$ and $A\nsubseteq V$.
Is it clearer now, how you should conclude?