Is really $[0,1]\times[0,1)$ homeomorphic to $[0,1)\times[0,1)$ with the product topology? I got asked to show that $[0,1]\times[0,1)$ is homeomorphic to $[0,1)\times[0,1)$ with the product topology, but I'm having trouble understanding why that's not false. $[0,1]$ is obviously compact while $[0,1)$ is not. What am I not getting here?
 A: A quick warm-up: how can it be that $A\times B\cong C\times B$ even when $A\not\cong C$?
The simplest example is the following: consider $A=\{a\}$, $C=\{c, d\}$ with the discrete topology; and let $B=\mathbb{N}$ with the discrete topology. So $A$ and $C$ consist of one and two unrelated points, respectively, and $B$ is just a blob of unrelated points.
Then it's easy to check that $A\times B\cong C\times B$ - in fact, both are homeomorphic to $B$ itself! This just boils down to arithmetic with infinities: twice an infinite set is no larger than the original infinite set.

Now onto your question. As it happens, there is a nice intuitive picture of why these two spaces are the same:
Picture $A=[0, 1)\times [0, 1)$ as a square with the top and left edges bold (= part of the space), and the bottom and right edges dashed (= not part of the space); similarly, picture $B=[0, 1]\times [0, 1)$ as a square with the top, left, and right edges bold, and the bottom edge dashed.
By stretching these around, we can view each as a disc, with the left half of the boundary bold and the right part of the boundary dashed. So they are homeomorphic.

Note that of course this isn't yet a rigorous answer to your question, since I haven't exhibited a homeomorphism (or proved one exists), but rather just (hopefully!) made it plausible that there is one. It takes a little bit of work to find an explicit homeomorphism that behaves the way the picture I describe above does, but it's not hard and is a good exercise.
