If $A\subsetneq X, B\subsetneq Y$, X,Y are connected then $X\times Y- A\times B$ is connected Question is to prove that :
If $A\subsetneq X, B\subsetneq Y$, and  $X,Y$ are connected then $X\times Y- A\times B$ is connected.
I do not immediately see why this is true. 
So, I thought if i can easily see something less is true. i.e., if $X$ is connected and $A$ is a proper subset of $X$, Then $X-A$ is connected.
take $X=\mathbb{R}$ ,and for $a\neq b$, $A=[a,b]$ then $X-A=\mathbb{R}-[a,b]$
we have $\mathbb{R}-[a,b]=(-\infty,a)\cup (b,\infty)$ 
$(-\infty,a)\neq \emptyset$ and $(b,\infty)\neq \emptyset$ and $(-\infty,a)\cap (b,\infty)=\emptyset$
somehow i have to say that $(-\infty,a)$ and  $(b,\infty)$ are not open in $\mathbb{R}-[a,b]$
for $M\subset X$ we say that $U$ is open in $M$ if $U=V\cap M$ for some open set $V$ in $X$ 
Suppose $(-\infty,a)$ is open in $\mathbb{R}-[a,b]$ then
we would have $(-\infty,a)=\mathbb{R}-[a,b] \cap V$ for some open set $V$ in $\mathbb{R}$
but $(-\infty,a)=\mathbb{R}-[a,b] \cap (-\infty,a)$ and $ (-\infty,a)$ is open in $\mathbb{R}$ thus,  
$(-\infty,a)$ is open in $\mathbb{R}-[a,b]$  similarly, $(b,\infty)$ is open in $\mathbb{R}-[a,b]$
I do not see where did i go wrong??
Please help me to sort this out..
thank you.
P.S : Though  my original question is to prove $X\times Y- A\times B$ is connected. while writing this, i was thinking of trying some thing less which i thought could be easily seen and then try for original question... but, I got stuck with simple thing only... But I feel "$A\subsetneq X, B\subsetneq Y$, and  $X,Y$ are connected then $X\times Y- A\times B$ is connected." then $A\subsetneq X$ , $X$ is connected then $X-A$ should also be true... please let me know why is this false.. I mean it should be... right???
 A: The proposition that "given $A\subsetneq X$ where $X$ is connected, then $A-X$ is connected" is not true, as you shown, but it does not follow from or is required for the original proposition, witch is true. To see why, notice that there exists a point $(x, y)$ in $X\times Y−A\times B$ where $x$ is not in $A$ and $y$ is not in $B$, and remember that connected components are disjoint or equal (this does not immediately prove the statement, there are another couple of steps). it might help you intuition to imagine $X=Y=[0, 1]$ thus, $X\times Y$ can be seen as a square, now imagine different $A$ and $B$'s. 
Elaboration (edit):
let $x_1\in X-A$ and $y_1\in Y-A$ 
it is enough to show that any point $(x,y)\in X\times Y-A\times B$ is in the same connected component as $(x_1, y_1)$
let $(x,y)\in X\times Y-A\times B$, 
without loss of generality $x\notin A$ so, 
$\{x\}\times Y\subset X\times Y-A\times B$ 
$\{x\}$ is connected and $Y$ is connected so $\{x\}\times Y$ is connected and thus $(x, y)$ and $(x, y_1)$ belong to the same connected component.
in the same way $X\times \{y\}$ is connected and $(x, y_1)$ and $(x_1, y_1)$ belong to the same connected component.
This out line is a bit rough, and should be polished, but you can get the general idea.
Edit:
as a response to the P.S, your second proposition is not weaker then the original, to put it very unmathematically , in a 2 dimensional space you can go around an obstacle, cutting a segment out of a road will split it into separate parts, but cutting a rectangle out of a parking lot will live it connected(so long as the rectangle does not span the whole width or height of the lot, thus $A\subsetneq X$ and not $A\subset X$)...
