k-Cells are Connected I am studying real analysis from Baby Rudin, and while the book proves that real intervals are connected, it does not say anything regarding k-cells. I would expect them to also be connected, but do not now how to prove it. 
I was thinking that if I were able to prove that: $A\subset\mathbb{R^n}$ connected , $B\subset\mathbb{R^m}$ connected $\Rightarrow$ $A\times B$ is connected (as a subset of $A\subset\mathbb{R^{n+m}}$), then the theorem would follow by induction using that $[a,b]$ is connected. However, I'm not sure how to prove this.
Thanks in advance!
 A: For now, let's work in $\mathbb{R}^2$ and consider some $k$-cell $S = [a, b] \times [c, d]$.  
Suppose for contradiction that a separation existed for $S$.  That is, there eixsts sets $A$ and $B$ such that $S = A \cup B$ and $A \cap B = A \cap \overline{B} = \overline{A} \cap B = \emptyset$.  
Choose any points $x \in A$ and $y \in B$ and consider the line segment $L$ connecting them.  Certainly, $L \subset S$, and the separation above would induce a separation on $L$.  In particular:
$$L = (A \cap L) \cup (B \cap L)$$
And further, since $A$ and $B$ are separated:
$$(A \cap L) \cap (B \cap L) = (\overline{A \cap L}) \cap (B \cap L) = (A \cap L) \cap (\overline{B \cap L}) = \emptyset$$
This is a contradiction, since line segments are connected.

Now, this is taking for granted that line segments are connected, which should follow from segments in $\mathbb{R}$ being connected.  For one, there exists a continuous function $f:[0, 1] \rightarrow L$.  
Point being, it can be proven that, if $A$ is connected and $f$ is continuous, then $f(A)$ is also connected.  The unfortunate thing is that Rudin doesn't introduce continuity until Chapter 4!
At any rate, this argument should generalize to $k$-cells in $\mathbb{R}^n$.
A: If you strengthen your claim to path connectedness, your induction step will become easy as any two given points $(a_1, b_1), (a_2, b_2) \in \mathbb{R}^m \times \mathbb{R}^n$ can both be connected to  $(a_2, b_1)$ by assumption.
Since intervals are trivially path connected and path connectedness implies connectedness (which is easy to show knowing that intervals are connected) this prooves your statement.
A: Hint: 
Show that all convex sets are connected (very much a 'follow your nose' type of proof starting with the definition of convexity), then show that all k-cells are convex (which again, is a relatively 'follow your nose' type of proof).
-hope this helps!
