Union of a connected set and its accumulation point [closed]

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.

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closed as off-topic by 900 sit-ups a day, Miha Habič, Jyrki Lahtonen, le gâteau au fromage, Sami Ben RomdhaneJul 30 at 7:59

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homework is not a stand alone tag on this site. Please use another subject tag. BTW, what have you tried? –  user21436 Apr 20 '12 at 3:13
Do you mean to say $B=A\cup {p}$? –  Han Altae-Tran Apr 20 '12 at 3:14
@HanAltae-Tran. How would you define $A\cup p$ since $p$ is not a subset but an element of $X$? –  Thomas E. Apr 20 '12 at 4:22
The question before was written as $B=AU{p}$. Of course, we must only consider unions of sets. –  Han Altae-Tran Apr 20 '12 at 23:24

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 $A$.

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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 $A$?

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I see the tag edit has happened. I'll remove these no longer relevant comments. –  user21436 Apr 20 '12 at 3:40