This is question numebr 13 exercise 3.1 wayne patty topology.  Let $p$ a cut point of a connected space $(X, \mathcal{T})$ and suppose $A$ and $B$ form a separation  of $X-\{p\}$. Prove that $A \cup \{p\}$ is conected.  Attempt: I proved A and B are connected using contradiction. Now , For A$\bigcup ${p} to be connected I need to  only prove that  $Closure (A) \bigcup ${p} $\neq \phi$.  I assumed let p doesnot belongs to closure of A and trying to get some contradiction but I am unable to do so. Please help!
Let $p$ a cut point of a connected space $(X, \mathcal{T})$ and suppose $C$ and $D$ form a separation  of $X-\{p\}$, i.e $$X-\{p\}=C \cup D$$ Prove that $C \cup \{p\}$ is conected. I have this idea: Since $p$ is a cut point then $X-\{ p \}$ is not connected, and since $C$ and $D$ form a separation of $X- \{p \}$ then $X-\{p \} = C \cup D$, and also $C \cup D$ is not connected. And also $$C \cap \overline{D}= \overline{C} \cap D= \emptyset$$ I need an idea to continue please
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