Let $X$ and $Y$ be connected spaces with proper subset $A$ and $B$. Then prove $(X\times Y)-(A\times B)$ is connected. I am very new to topology and working my way through Munkres. I have seen a few proofs of this, but none the way I approached it. So I am wondering, is the following correct? If not, where is my logic flaw?
Note $X$ and $Y$ connected implies $X\times Y$ is connected.  Now assume to the contrary that $(X\times Y)-(A\times B)$ is disconnected.
Let $C_1\times C_2$ and $D_1\times D_2$ be the separation.  Thus $C_1\times C_2$ and $D_1\times D_2$ are disjoint, open and nonempty so that $(C_1\times C_2)\cup(D_1\times D_2)=(X-A)\times (Y-B)$.  For $C_1\times C_2$ to be disjoint from $D_1\times D_2$ then either $C_1$ and $D_1$ are disjoint or $C_2$ and $D_2$ are disjoint.
If $C_1$ and $D_1$ are disjoint, then these sets form a separation of $X-A$ ($C_1, D_1$ are open, disjoint and nonempty in $X-A$).  Thus $C_1$ and $D_1\cup A$ are disjoint, open and nonempty.  Therefore we have a separation of $X$.  A contradiction. 
Similar logic applies to $C_2\times D_2$ disjoint. 
 A: As $X$ and $Y$ are connected, every $Y_x = \{x\} \times Y$ is also connected ($x \in X$) and also all $X_y:= X \times \{y\}$ ($y \in Y$) are connected, as homeomorphic copies of $Y$ and $X$ resp.
Now fix $x_0 \in X-A$, which can be done as $A$ is a proper subset of $X$.
Also fix $y_0 \in Y-B$, likewise.
Define $$C = Y_{x_0} \cup \bigcup_{y \in Y-B} X_y \subseteq (X \times Y) - (A \times B)$$
and symmetrically $$D= X_{y_0} \cup \bigcup_{x \in X-A} Y_x \subseteq (X \times Y) - (A \times B)$$
Both $C$ and $D$ are connected as all sets in the union are connected and for $C$ each $X_y$ intersects $Y_{x_0}$ (in $(x_0,y)$) and for $D$ each $Y_x$ intersects $X_{y_0}$ (in $(x,y_0)$) and standard theorems on unions of intersecting connected sets will guarantee that $C$ and $D$ are both connected and it's easy to check that $$C \cup D = (X\times Y)-(A\times B)$$ and $(x_0,y_0) \in C \cap D$, so again $C \cup D$, and hence, $(X\times Y)-(A\times B)$ is connected.
A: Here is a proof using only the definition of connectedness.  Suppose that $S:=(X\times Y)\setminus (A\times B)$ is a union of two disjoint open subsets $U$ and $V$.  We shall show that either $U=\emptyset$ or $V=\emptyset$.
For each $x\in X$, let $Y_x:=\{x\}\times Y$.  For each $y\in Y$, let $X^y:=X\times\{y\}$.  Note that $Y_x\cong Y$ and $X^y\cong X$ for all $x\in X$ and $y\in Y$.  Therefore, both $Y_x$ and $X^y$ are connected sets.
Since $A$ and $B$ are proper subsets of $X$ and $Y$ respectively, we can choose $u\in X\setminus A$ and $v\in Y\setminus B$.  Without loss of generality, we assume that $(u,v)\in U$.  
Because $Y_u\cap U$ and $Y_u\cap V$ are disjoint open subsets of $Y_u$ whose union is the whole $Y_u$, and $Y_u\cong Y$ is connected, we deduce that $Y_u\cap U=\emptyset$ or $Y_u\cap V=\emptyset$.  As $(u,v)\in Y_u\cap U$, we must have $Y_u\subseteq U$ and $Y_u\cap V=\emptyset$.  Similarly, $X^v\subseteq U$ and $X^v\cap V=\emptyset$.
Let now $(x,y)\in S$ be arbitrary.  Then, either $x\notin A$ or $y\notin B$.  Suppose first that $x\notin A$.  Using the same argument as before, either $Y_x\cap U=\emptyset$ or $Y_x\cap V=\emptyset$.  However, $Y_x\cap U$ contains $Y_x\cap X^v$, which contains $(x,v)$.  This means $Y_x\cap U$ is nonempty, and so $Y_x\subseteq U$.  Thus, $(x,y)\in U$.  Similarly, if $y\notin B$, then we also obtain $(x,y)\in U$.
Consequently, every point of $S$ is in $U$.  Therefore, $U=S$, and $V$ must be empty.  This shows that $S$ is connected.
