Is this proof regarding product of connected spaces correct? Let $X,Y$ be connected spaces, and consider their product $X\times Y$. I want to show that their product is connected. The posts I've read here regarding this question often include creating "slices" $X\times a$ and $x\times Y$ for some point $a\times b \in X\times Y$ and showing that the union of the "slices" are connected. I tried a different approach, and my "proof" was very short so I was concerned whether or not I had made some grave mistakes. Here it is, and please be gentle with me.
So I assume that $X\times Y$ is not connected, and hopefully it will imply that either $X$, $Y$ or both cannot be connected spaces. Suppose that $(A\times B)\cup (C\times D) = X\times Y$ is a separation of $X\times Y$. Then $(A\times B) \cap (C\times D) = \emptyset$, implying that $A\cap C = \emptyset = B\cap D$. Also, from our separation we have that $A\cup C = X$ and $B\cup D = Y$, which are separations of both $X$ and $Y$, and this is our contradiction.
Is this wrong? It seems so much simpler than the other proofs I've seen, but I cannot fathom how if this is correct, nobody has mentioned it. So, where is the mistake guys?
 A: Your "grave mistake" is, as Tobias Kildetoft remarked, that you are assuming too special form
for open subsets of $X\times Y$. I tried your approach, starting with a split of the product space
into two open subsets (which are not of the special form assumed by you) and got through.
Again the slices play an important role, although we see them acting from our seats behind the scenes,
as it were.
So, let us assume that $X\times Y=U\cup V$, where $U$ and $V$ are disjoint open subsets of $X\times Y$;
we have to prove that one of the sets $U$, $V$ is empty.
For each $x\in X$ we define the continuous map $\varphi_x\colon Y\to X\times Y : y\mapsto (x,y)$.
Since the connected space $Y$ is a disjoint union of the open subsets $\varphi_x^{-1}(U)$
and $\varphi_x^{-1}(V)$, we have either $\varphi_x^{-1}(U)=Y$ and $\varphi_x^{-1}(V)=\varnothing$,
or $\varphi_x^{-1}(U)=\varnothing$ and $\varphi_x^{-1}(V)=Y$. Let $A:=\{x\in X\mid \varphi_x^{-1}(U)=Y\}$;
then $\{x\in X\mid \varphi_x^{-1}(V)=Y\}=A^{\mathrm{c}}$ ($=X\setminus A$). We have $U=A\times Y$
and $V=A^{\mathrm{c}}\times Y$. The projection $p\colon X\times Y\to X$ is an open map: it maps open sets
to open sets. But then $A=p(U)$ and $A^{\mathrm{c}}=p(V)$ are open subsets of the connected
space $X$, thus either $A$ is empty and hence $U$ is empty, or $A^{\mathrm{c}}$ is empty
and hence $V$ is empty.
[Added later.] $~$Let us be more ambitious.

Let $I$ be a set, finite or infinite; we are assuming that $I$ has at least two elements, to avoid trivialities. For each $i\in I$ let $X_i$ be a connected
  topological space; we are assuming that
  all $X_i$ are nonempty, again to avoid trivialities.
We claim that the product space $Y := \prod_{i\in I} X_i$ is connected.

Proof. $~$Supose that $Y$ is a disjoint union of two open subsets $U$ and $V$; we have to prove that
one of the sets $U$, $V$ is empty. We shall prove the following equivalent assertion:
if $U$ is nonempty,
then $U=Y$. (Think about this for a moment or two: assume that the alternative
assertion is true; then either $U$ is empty, or $U$ is nonempty hence $U=Y$ and $V$ is empty.)
But for starters we assume nothing about emptiness/nonemptiness of $U$ or $V$.
Let $j\in I$, and set $Z_j\:=\prod_{i\in I\setminus\{j\}}\! X_i$.
For any $z\in Z_j$ we define the continuous mapping $\varphi_z\colon X_j\to Y$,
where for
every $x\in X_j$ the point $\varphi_z(x)\in Y$ is defined by
$(\varphi_z(x))_i=z_i$ if $i\in I\setminus\{j\}$, and $(\varphi_z(x))_j=x$.
The connected space $X_j$ is the union of two disjoint open sets
$\varphi_z^{-1}(U)$ and $\varphi_z^{-1}(V)$, therefore
either $\varphi_z^{-1}(U)=X_j$ and $\varphi_z^{-1}(V)=\varnothing$,
or $\varphi_z^{-1}(U)=\varnothing$ and $\varphi_z^{-1}(V)=X_j$.
This conclusion holds for every $j\in I$ and every $z\in Z_j$.
We can rephrase it as follows:
if two points $y$ and $y'$ of $Y$ differ
in at most a single component
$($that is, there is $j\in I$ such that $y_i=y'_i$ for all $i\in I\setminus\{j\}$$)$,
then either both belong to the open set $U$ or both belong to the open set $V$.
Using a simple induction we further conclude that any two points of $Y$ that differ in finitely
many components either both lie
in $U$ or both lie in $V$.
Now we do assume that $U$ is nonempty, and choose a point $u\in U$.
There exists a basic open set $B$ of the product space $Y$
which contains the point $u$ and is contained in the open set $U$.
This means that $B=\prod_{i\in I} B_i$, where $B_i$ is an open set in $X_i$ for each $i\in I$, and there exists a finite subset $K$
of $I$ such that $B_i=X_i$ for every $i\in I\setminus K$; moreover, $u_i\in B_i$ for every $i\in I$,
and $B\subseteq U$. Let $y$ be any point in $Y$; we claim that $y$ lies in $U$.
$($In other words, we claim that $Y=U$, so we are done as soon as we prove it.$)\,$ We construct the point $z\in B\subseteq U$ as follows:
$z_i=y_i$ for $i\in I\setminus K$, and $z_k=u_k$ for $k\in K$.
The point $y$ differs from the point $z\in U$ only in finitely many components,
therefore $y\in U$.$~$ Done.
A: There is a very simple reason why the product is connected, based on your idea of slices.  
Remark: Every slice $\{x\}\times Y$ is connected as it is a copy of $Y$. Similarly, every slice $X\times \{y\}$ is connected because it is a copy of $X$. 
Suppose $A$ is nonempty and clopen in the product $X\times Y$.
Let $(a,b)\in A$. By the the first part of the remark, $\{a\}\times Y\subseteq A$. So every horizontal slice meets $A$. Now by the second part of the remark every horizontal slice is contained in $A$. Thus $A=X\times Y$.
