Suppose $Y\subset X$ and $X,Y$ are connected and $A,B$ form separation for $X-Y$ then, prove that $Y\cup A$ and $Y\cup B$ are connected Question is :
Suppose $Y\subset X$ and $X,Y$ are connected and $A,B$ form separation for $X-Y$ then, Prove that  $Y\cup A$ and $Y\cup B$ are connected.
What i have tried is : 
Suppose $Y\cup A$ has  separation say $C\cup D$ then all i could see is that 
Either $Y\subset C$ or $Y\subset D$ as $Y$ is connected and $
C\cup D$ is separation for a subset that contains $Y$.
Without loss of generality we could assume $Y\subset C$
As $Y\cup A=C\cup D$ and $Y\subset C$  i can say $D\subset A$ (I do not know how does this help)
I have $X-Y=A\cup B$ with $A\cap B=\emptyset$
I have not used connectedness of $X$ till now, So i thought of using that and end up with following : 
$X-Y=A\cup B\Rightarrow X=Y\cup A\cup B=(Y\cup A)\cup (Y\cup B)$
I do not know what to conclude from this... 
I would be thankful if some one can help me by giving some "hints"
Thank you
 A: As $A,B$ form a separation for $X-Y$ and $X$ is connected, $Y$ must be non-empty.
You suppose that $Y\cup A$ has a separation $C\cup D$. Then $Y$, being connected, is a subset of $C$
Now $A$ is clopen in $X-Y$, that means there are sets $U$ open and $K$ closed, such that $U-Y=K-Y=A$. You know that $D\subseteq A$ is clopen in $Y\cup A$. Can you show that $D$ is open in $U$ and closed in $K$?
A: Continuing with your argument, We can prove that $B \cup C$ and $D$ form the partition of $X$.(Assuming your conclusion that $D \subset A$). 
We have $X = (B \cup C) \cup D$.
$C$ and $D$ form separation of $Y \cup A$. So, using theorem (23.1), no limit point of $C$ is in $D$. Similarly, no limit point of $B$ is in $D$ since $D \subset A$ and $A$ and $B$ form separation of $X-Y$. From this, we can easily say that no limit point of $B \cup C$ is in $D$ i.e. Every limit point of $B \cup C$ wrt $X$ is in $B \cup C$. So, $B \cup C$ is closed in $X$ and $D$ is open in $X$.
Now, use a similar argument that, no limit point of $D$ is  in $B \cup C$ implying that $D$ is closed in $X$ and $B\cup C$ is open in X. Hence, $B \cup C$ and $D$ are $\textit{clopen}$ in $X$ and form partition of $X$.
A: We have $X \setminus Y = A \cup B$. On the other hand as $Y \subset X$, then $Y \cup (X \setminus Y) = X \cup Y = X$. Hence, of the equality $Y \cup (X \setminus Y) = Y \cup (A \cup B)$, we have that $X = (Y \cup A) \cup (Y \cup B)$. And since $X$ is connected, this last union is connected.
You can apply a theorem that says the following: the union of a collection of connected subspaces of $X$ that have a point in common is connected.
A: Let $A',B'$ be open in $X$ with $A'\cap (A\cup B)=A$ and $B'\cap (A\cup B)=B.$ [ So $A'\cap B'\subset Y.$ ]
Suppose $Y\cup A\subset C'\cup D'$ where $C',D'$ are open in X, with $C'\cap (Y\cup A),\; D'\cap (Y\cup A)$ being disjoint.$\quad$ [ So $D'\cap C'\subset B$ . ]
Then $C'\cap Y,\, D'\cap Y$ are open in $Y$ and disjoint, and cover $Y.$ Since $Y$ is connected, one of them, say $D'\cap Y,$ is empty.
So $Y\subset C'.$ We show that $Y\cup A$ is connected by showing that $A$ is also a subset of $C',$ as follows:
Let $E'=A'\cap D'$ and let $F'=C'\cup B'.$
We have  $E'\cap C'=A'\cap (D'\cap C')\subset A'\cap B=\emptyset.$
We have $E'\cap B'=D'\cap (A'\cap B')\subset  D'\cap Y=\emptyset.$
Therefore $E',F'$ are disjoint. 
We have $Y\cup B\subset C'\cup B\subset C'\cup B'=F'.$
We have $A=(A\cap D')\cup (A\cap C')\subset (A'\cap D')\cup (A\cap C')=E'\cup (A\cap C')\subset E'\cup F'.$ 
Therefore, since $X=(Y\cup B)\cup A,$ we have  $E'\cup F'=X.$
Now we have $E'=\emptyset$ because (i) $E',F'$ are disjoint and open in $X$ with $E'\cup F'=X,$ and $X$ is connected, so one of $E', F'$ is empty, and (ii) $Y\subset F'$ and $Y$ is not empty, otherwise $X=A\cup B=A'\cup B'$ would be a disconnection of $X.$
Hence $X=F'.$ So $A=A\cap F'=A\cap (C'\cup B')=(A\cap C')\cup (A\cap B')=(A\cap C')\cup \emptyset \subset C'$ as desired.
