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

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 $X$.
