Why are all k-cells convex? Reading through the first half of Baby Rudin again before taking an Analysis class, I came across the assertion that "it is also easy to show that k-cells are convex".  
Previously it gave the example of open/closed balls being convex, and the proof is obvious and easy to understand.  That being said, and it's probably very easy, but I can't for the life of me produce something that shows that all k-cells are convex as well.
I understand convex to sort of say "all points 'between' two points in a given set are also in the set" in Layman's terms, but I'm not sure where to proceed from there.  I either got this on my first read and can't remember for the life of me, or I took it for granted and went with it.  
Any help leading me in the right direction?

Suppose that k-cell means the product of intervals in $\Bbb R^k$. Let $C=\prod_{n=1}^k I_n $, where $I_n$ is an interval for each $n$, be a k-cell. Recall that in $\Bbb R$ $I$ is an interval iff for $a,b\in I$ such that $a<b$, $c\in (a,b)$ implies that $c\in I$.
Now let $x,y\in C$ where $x=(x_1,x_2,...,x_k)$ and $y=(y_1,y_2,...,y_k)$. Any convex combination of $x$ and $y$ is if the form $z=tx+(1-t)y$, $t\in [0,1]$. Therefore
$$ z=(tx_1+(1-t)y_1,tx_2+(1-t)y_2,...,tx_k+(1-t)y_k)$$
It is easy to see that each component of $z$ belongs to its respective $I_n$ since $tx_n+(1-t)y_n$ is between $x_n$ and $y_n$. Thus $z$ belong to $C$ as claimed.
