Union of a connected set and its accumulation point

Question

Let $A$ be a connected set in the metric space $(X, d)$. If $p$ is an accumulation point of $A$,then prove that $B = A \cup \{p\}$ is connected.

Answer 1

Suppose the contrary: there exists a separation $B_{1},B_{1}$ of $U:=A\cup \{p\}$. Wlog we may assume that $p\in B_{1}$. If $A\cap B_{1}\neq\emptyset$ then we have a contradiction since $A$ is connected, hence $A\cap B_{1}=\emptyset$. Since $B_{1}$ is open in $U$ there exists $r>0$ so that $B_{U}(p,r)\subset B_{1}$, which is a neighborhood of $p$ that does not intersect $A$. This is a contradiction since $p$ was an accumulation point of $A$.

Answer 2

Suppose you have a separation. Then $p$ is contained in a nontrivial clopen set. What else can you say about this set? What can you conclude about $A$?