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

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.
