Htykuut

I have been asked to prove the following:

Let $X$ be a connected topological space, let $pin X$, and let $X-{p}$ be disconnected. Prove that ${p}$ must be either open or closed in $X$, but cannot be both open and closed in $X$.

I understand that ${p}$ cannot be both open and closed in $X$ as this would imply that ${p}$ and $X-{p}$ are both open in $X$, which implies that $(X-{p}) cup{p}=X$ is disconnected, as it is the union of two open disjoint sets.

So far I have been unable to establish why the case of ${p}$ being neither closed nor open is untenable. If you have any input or suggestions, it would, as always, be greatly appreciated.

Since $Xsetminus{p}$ is disconnected, there are open non-empty subsets $A$ and $B$ of $Xsetminus{p}$ whose intersection is empty and whose union is $Xsetminus{p}$. Since $A$ and $B$ are open subsets of $Xsetminus{p}$, there are open subsets $A^star$ and $B^star$ of $X$ such that $A=A^starcap(Xsetminus{p})$ and that $B=B^starcap(Xsetminus{p})$. Now, there are $4$ possibilities:

1. 
   $pnotin A^star$ and $pnotin B^star$: then $A=A^star$ and $B=B^star$. So, ${p}$ is a closed set, since $Xsetminus{p}=A^starcup B^star$, which is an open set.


2. 
   $pin A^star$ and $pnotin B^star$: this is impossible, because then $X$ would be disconnected, since $A^starcap B^star=Acap B=emptyset$, $A^starcup B^star=X$, and $A^star,B^starneqemptyset$.


3. 
   $pnotin A^star$ and $pin B^star$: its like the previous case.


4. 
   $pin A^star$ and $pin B^star$: then ${p}=A^starcap B^star$ and therefore ${p}$ is an open subset of $X$.