Is is possible that $A \times C$ homeomorphic to $B \times C$? 
Suppose that $A$ and $B$ are non-homeomorphic subsets of $\mathbb{R}$ and $C$ is a subset of $\mathbb{R}$. Is it possible that $A \times C$ is homeomorphic to $B \times C$?

I'm not even sure if its possible or not, so I have no idea what a counterexample would even be. Any ideas? 
 A: $[0,1) \times [0,1)$ is homeomorphic to $[0,1] \times [0,1)$ and $[0,1)$ is not homeomorphic to $[0,1]$. So it is indeed possible with subsets of $\mathbb{R}$.
A: Yes — as Adayah has pointed out,
If $A = \{0\}$ is a singleton set, and $B = C = \mathfrak{C}$, the Cantor set, then
$$A \not\cong B \qquad A\times C =\mathfrak{C}  \qquad B\times C = \mathfrak{C}^2$$
and $\mathfrak{C}$ is homeomorphic to $\mathfrak{C}^2$ via the map that “unzips” the ternary representation:
$$\langle a_1b_1a_2b_2 \ldots \rangle \mapsto \langle a_1a_2a_3\ldots\rangle\times \langle b_1b_2b_3\ldots\rangle $$
A: Is it possible? Yes. 

Let $A=(0,1),B=[0,1]$ and $C=\emptyset$. Then $A$ and $B$ are not homeomorphic as $A $ is not compact, but $B$ is. Further, $$A \times C=B\times C$$Thus, $A \times C$ is homeomphic to $B \times C.$

Is it always true? No.

Let $A=\{0\},B=\mathbb R$ and $C=\mathbb R.$ Then Then $A$ and $B$ are not homeomorphic as $B$ is not compact, but $A$ is.
  Further, $A \times C$ is homeomorphic to $\mathbb R$ and $B \times C$ is homeomorphic to $\mathbb R^2.$

