If $X$ and $Y$ are connected, then $(X\times Y)\setminus(A\times B)$ is connected for any proper subsets $A,B$ I meet these two exercises:  


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*Q1: let $A$ be a proper subset of $X$, and $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, show that $(X\times Y)\setminus(A\times B)$ is connected.  

*Q2: Let $Y\subset X$. Assume that $X$ and $Y$ be connected. Show that if $A$ and $B$ form a separation of $X\setminus Y$, then $Y\cup A$ and $Y\cup B$ are connected.   
My attempt for Q2
I think to prove it by contradiction, assume $Y\cup A$ and $Y\cup B$ are not connected
then for $P$ and $Q$ disjoint $Y\cup A=P\coprod Q$ and for $M$ and $N$ disjoint $Y\cup B=M\coprod N$
$$(Y\cup A)\cup (Y\cup B)=(P\coprod Q)\cup (M\coprod N)$$
The left side will give $X$,and the right side can be written as a disjoint union , this contradicts the fact that $X$ is connected, so $Y\cup A$ and $Y\cup B$ must be connected.
I need help for Q1.  
 A: As to Q1, for every $x$ in $X \setminus A$, $S_{x} = \{x\} \times Y$ is connected,
and for every $y$ in $Y \setminus B$, $T_{y} = X \times \{y\}$ is connected, as these sets are homeomorphic to the connected spaces $Y$ resp. $X$. Also, every $S_{x}$ intersects $T_{y}$, so
for fixed $x \in X \setminus A$, $U_{x} = S_{x} \cup \cup_{y \in Y \setminus B} T_{y}$ is connected as well.
Now $( X \times Y) \setminus (A \times B)$ is the union of these sets $U_x$, which all intersect as well, making it connected. This uses standard theorems on unions of connected sets.
As to Q2, it suffices to show (by symmetry) that $Y \cup A$ is connected. Suppose not, then it can be written as a disjoint clopen union (non-trivially) of say $U$ and $V$. 
By connectedness of $Y$ one of them, say $U$ must miss $Y$, and then one checks that $U$ is clopen in $A$. As $A$ is clopen in $X \setminus Y$, $U$ would then be clopen in $X \setminus Y$ and thus in $X$, contradicting its connectedness. Some details omitted.
A: In question 2 you have a logical mistake in the assumption: if you prove it by contradiction then you have to assume that one of them is disconnected, not necessarily both. Also, you need to mention that $P,Q$ are open (or closed).
As for Q1, prove by contradiction: If $(X\times Y)\setminus(A\times B)$ is disconnected then $$(X\times Y)\setminus(A\times B)=F_1\times F_2 ⊔ G_1\times G_2$$
$F_1,G_1$ are closed in $X$, $F_2,G_2$ are closed in $Y$. Since $A$ is proper, take any $x\notin A$. Then for any $y\in Y$ you have $(x,y)\in(X\times Y)\setminus(A\times B)$, hence $(x,y)\in F_1\times F_2 ⊔ G_1\times G_2$. It follows that $Y=F_2 ⊔G_2$ (Fill in all the details)
