Product of connected metric spaces I want to show that if $X$ and $Y$ are connected metric spaces, then $X \times Y$ is also connected. As a hint it is marked to write $X \times Y$ as Unions of sets of the form $(X \times \{y\}) \cup ( \{x\} \times Y)$, but i don't have an idea how to continue. Any hints?
 A: HINT: Try it with $[0,1]\times [0,1]$ first. How can you apply the hint here? Once you can do this, the general case should follow fairly quickly.
A: Another different proof from wiki:
Let us begin with a Lemma:

If a topological space $X$ is connected iff for any point $x \in X$ and for any open cover $U$ of $X$, we have $\cup^\omega_{n=1}St^n(x,U)=\cup \{u \in U: x\in u\}=X$, where $St^1(x,U)=\cup \{u \in U: x\in u\}$, and when $n>1$, $St^n(x,U)=St(St^{n-1}(x,U)).$

Proof: 


*

*Left to right: We first assume that $X$ is connected. For any open cover $U$ of $X$, and for any $x\in X$, if there is a point $y$ such that $y$ is not in  $\cup^\omega_{n=1}St^n(x,U)$, therefore there exists open set family $V \subset U$ $y$ such that $\cup V \cap \cup^\omega_{n=1}St^n(x,U)= \emptyset$, which implies $\cup V$ is an nonempty open-closed subspace of $X$, contradiction! 

*Right to left: if the space is not connected, then exists an nonempty closed-open subspace $Y$. We let the open cover $Y\cup (X-Y)$, obviously, it can't satisfies the condition.


Then the proof for the question is not difficult, I believe. Try. (Drawing a picture will be more clear:) 
A: As for each $x\in X$ the subspace $\{x\}\times Y$ is homeomorphic to the connected space $Y$ (via the projection), it is connected. The same holds for all $X×\{y\}$, $y\in Y$. Now if $(x,y)$ and $(x',y')$ are points in $X×Y$, then
$$(\{x\}×Y) \cup (X×\{y'\})$$
is connected since it's intersection $\{(x,y')\}$ is non-empty. This shows that there is a connected set containing $(x,y)$ and $(x',y')$. Since these points were arbitrary, $X×Y$ is connected.
