Prob. 9, Sec. 23, in Munkres' TOPOLOGY, 2nd ed: If $X$ and $Y$ are connected and if $A$ and $B$ are proper subsets of $X$ and $Y$, resp., then Here is Prob. 9, Sec. 23, in the book Topology by James R. Munkres, 2nd edition: 

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

My Attempt: 

It can be shown that 
  $$ (X \times Y) \setminus (A \times B) = [ X \times (Y \setminus B) ] \cup [ (X \setminus A ) \times Y ]. $$
  Let us choose a point $u \in X \setminus A$ and a point $v \in Y \setminus B$. 
The space $u \times Y$, being homeomorphic with the connected space $Y$, is connected, and so is the space $X \times y$ for any point $y \in Y$ and hence for any point  $y \in Y \setminus B$. 
Now, let us take any point $y \in Y \setminus B$. As the spaces $u \times Y$ and $X \times y$ are connected and as these have the point $u \times y$ in common, so the union $ ( u \times Y ) \cup (X \times y)$ is also connected, by virtue of Theorem 23.3 in Munkres. 
Moreover,  the space $X \times v$, being homeomorphic with the connected space $X$, is connected, and so is the space $x \times Y$ for any point $x \in X$ and hence for any point $x \in X \setminus A$. 
Now, let us take any point $x \in X \setminus A$. As the spaces $x \times Y$ and $X \times v$ are connected and as they have the point $x \times v$ in common, so their union $( x \times Y) \cup ( X \times v)$ is also connected. 
Now, as, for each point $x \in X \setminus A$ and for each point $y \in Y \setminus B$,  the spaces $ ( u \times Y ) \cup (X \times y)$ and $( x \times Y) \cup ( X \times v)$ are connected and as they have the point $u \times v$ in common, so the union 
  $$ [ ( u \times Y ) \cup (X \times y) ] \cup [ ( x \times Y) \cup ( X \times v) ] $$
  is also connected. 
Finally, as, for each point $x \in X \setminus A$ and for each point $y \in Y \setminus B$, the spaces 
  $$ [ ( u \times Y ) \cup (X \times y) ] \cup [ ( x \times Y) \cup ( X \times v) ] $$
  are connected and as these have the point $u \times v$ in common, so the union 
  $$ \bigcup_{ x \in X\setminus A, \ y \in Y \setminus B} \left[ \  [ ( u \times Y ) \cup (X \times y) ] \cup [ ( x \times Y) \cup ( X \times v) ]  \ \right] $$ 
  is also connected, and this set coincides with $( X \times Y) \setminus ( A \times B)$. 

Am I right? 
Is this proof correct? If so, then is my presentation good enough too? Have I introduced any redundancies in my reasoning? 
If this proof (or portions thereof) is not correct, then where lie the problems? 
 A: I would like to start with a generalization of theorem 23.3 that makes things more easy.

Generalization of theorem 23.3: If $\left\{ A_{\alpha}\right\} $
is a collection of subspaces of $X$ such that every $A_{\alpha}$
is connected and such that for every pair $\left\langle \alpha,\beta\right\rangle $
the intersection $A_{\alpha}\cap A_{\beta}$ is not empty, then the
union of that collection is connected.
Proof: let $Y$ denote the union and suppose that $Y=C\cup D$ is
a separation of $Y$. Since $A_{\alpha}$ is connected it must be
a subset of $C$ or $D$. WLOG we may assume that $A_{\alpha_{0}}\subseteq C$
for some $\alpha_{0}$. Then the condition that $A_{\alpha_{0}}\cap A_{\alpha}\neq\varnothing$
leads to the conclusion that also $A_{\alpha}\subseteq C$ for every
$\alpha$. So we find that $Y\subseteq C$ or $Y\subseteq D$, which
contradicts that $C\cup D$ is a separation of $Y$. This shows that
no separation of $Y$ exists.

Note that in the generalization it is not demanded that the collection has a common point but only that every pair has a common point. 
Now I go on with a solution of problem 9 based on that generalization.

For $x\in X$ and $y\in Y$ define $C_{x,y}:=\left(\left\{ x\right\} \times Y\right)\cup\left(X\times\left\{ y\right\} \right)\subseteq X\times Y$.
So if $X$ and $Y$ are connected then $C_{x,y}$ is the union of
two connected sets with non-empty intersection, hence $C_{x,y}$ is
connected. Further we have $C_{x,y}\cap C_{x',y'}\neq\varnothing$
because it contains $\left\langle x,y'\right\rangle $ and $\left\langle x',y\right\rangle $
as elements. Now observe that: $$\left(X\times Y\right)-\left(A\times B\right)=\bigcup_{x\in X-A,y\in Y-B}C_{x,y}$$This especially because  $A$ is proper subset of $X$ and $B$ is a proper subset of $Y$. Then applying the generalization of 23.3 we conclude that $\left(X\times Y\right)-\left(A\times B\right)$
is connected.

I haven't checked your proof yet, but soon I will do that and give you an account of that.

edit:
I have checked your proof. It is completely okay and does not contain essential redundancies. 
